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Punjab Board Class 11 Mathematics

Chapter 8: Mathematical Induction and Binomial Theorem — Class 11 Maths Notes

Mathematical induction proofs with strict structure, plus binomial expansion: general term, middle term and approximations. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 8.

Exercise solutions — Chapter 8

Every exercise of Mathematical Induction and Binomial Theorem solved step by step.

Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4

Mathematics 11 (Punjab Board)

Chapter 8: Mathematical Induction and Binomial Theorem — Theory

This chapter has two big ideas. First, mathematical induction: a method to prove that a statement is true for every positive integer. Second, the binomial theorem: a formula that expands powers like \((a+b)^n\) without multiplying everything out by hand.

8.1 Principle of Mathematical Induction

Suppose we want to prove a statement \(S(n)\) for every positive integer \(n\). The principle of mathematical induction says it is enough to check two conditions:

Step 1 — Base case: Show that \(S(1)\) is true.

Step 2 — Inductive step: Assume \(S(k)\) is true for some positive integer \(k\) (this assumption is called the induction hypothesis), and then prove that \(S(k+1)\) is also true.

If both steps hold, then \(S(n)\) is true for every positive integer \(n\). Is ka matlab hai: pehli seedhi (base case) par kharay ho jao, aur phir sabit karo ke har seedhi se agli seedhi par charha ja sakta hai — to tum har seedhi tak pahunch sakte ho.

Key point: In the inductive step you must use the induction hypothesis \(S(k)\) to prove \(S(k+1)\); a proof that never uses it is not induction.

8.1.1 Extended Principle of Mathematical Induction

Sometimes a statement is only claimed for integers \(n \ge n_0\) where \(n_0\) is not 1 (for example \(n \ge 4\), or even \(n \ge -2\)). The extended principle works the same way: prove the base case \(S(n_0)\), assume \(S(k)\) for \(k \ge n_0\), and prove \(S(k+1)\).

Key point: The base case must match the claim. If the claim starts at \(n = 4\), checking \(n = 1\) proves nothing.

8.2 Binomial Theorem (Positive Integral Index)

For any positive integer \(n\) and any real numbers \(a, b\):

\[ (a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{n}b^n = \sum_{r=0}^{n}\binom{n}{r}a^{\,n-r}b^{\,r} \]

The numbers \(\binom{n}{r} = \dfrac{n!}{r!(n-r)!}\) are called binomial coefficients. The expansion has \(n+1\) terms. Notice the pattern: as \(r\) increases, the power of \(a\) falls and the power of \(b\) rises, and in every term the two powers add up to \(n\).

The general term (the \((r+1)\)-th term) is:

\[ T_{r+1} = \binom{n}{r}a^{\,n-r}b^{\,r}, \qquad 0 \le r \le n \]

Key point: \(T_{r+1}\) means the term with index \(r\); for the 6th term put \(r = 5\). Students often lose marks by putting \(r = 6\).

8.2.1 Middle Terms in the Expansion of \((a+b)^n\)

If \(n\) is even, there is one middle term: \(T_{\frac{n}{2}+1}\). If \(n\) is odd, there are two middle terms: \(T_{\frac{n+1}{2}}\) and \(T_{\frac{n+3}{2}}\).

Also useful: \((a+b)^n + (a-b)^n\) keeps only the even-powered terms of \(b\) (the odd ones cancel), while \((a+b)^n - (a-b)^n\) keeps only the odd-powered terms.

\[ (a+b)^n + (a-b)^n = 2\left[\binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \binom{n}{4}a^{n-4}b^4 + \dots\right] \]

8.2.2 Deductions from the Binomial Expansion

Putting \(a = 1, b = x\) gives \((1+x)^n = \sum_{r=0}^{n}\binom{n}{r}x^r\). Two quick substitutions give famous identities:

\[ x = 1: \quad \binom{n}{0}+\binom{n}{1}+\dots+\binom{n}{n} = 2^n \]
\[ x = -1: \quad \binom{n}{0}-\binom{n}{1}+\dots+(-1)^n\binom{n}{n} = 0 \]

Also \(r\binom{n}{r} = n\binom{n-1}{r-1}\), which gives \(\binom{n}{1}+2\binom{n}{2}+\dots+n\binom{n}{n} = n\cdot 2^{n-1}\).

Key point: Whenever a question asks for a sum of binomial coefficients, try substituting a clever value of \(x\) into \((1+x)^n\).

8.3 Binomial Theorem for Negative or Fractional Index

When \(n\) is a negative integer or a fraction, the expansion of \((1+x)^n\) never stops — it is an infinite series, valid for \(|x| \lt 1\):

\[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots, \qquad |x| \lt 1 \]

The general term is \(T_{r+1} = \dfrac{n(n-1)\dots(n-r+1)}{r!}x^r\). Common special cases:

\[ (1+x)^{-1} = 1 - x + x^2 - x^3 + \dots, \qquad (1-x)^{-1} = 1 + x + x^2 + x^3 + \dots \]
\[ (1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \dots, \qquad (1+x)^{\frac{1}{2}} = 1 + \frac{x}{2} - \frac{x^2}{8} + \dots \]

To expand \((a+b)^n\) with this index, first take out the constant part: \((a+b)^n = a^n\left(1+\frac{b}{a}\right)^n\), then expand with \(x = b/a\) (needs \(|b| \lt |a|\)).

Key point: For small \(x\), higher powers can be dropped: \((1+x)^n \approx 1 + nx\) (neglecting \(x^2\) and above). This is the engine behind all the approximation questions.

8.4 Pascal's Triangle

Pascal's triangle lists the binomial coefficients row by row. Each entry is the sum of the two entries above it, which is the identity \(\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}\). Row \(n\) gives the coefficients of \((a+b)^n\):

\[ \begin{array}{c} n=0: \quad 1 \\ n=1: \quad 1 \quad 1 \\ n=2: \quad 1 \quad 2 \quad 1 \\ n=3: \quad 1 \quad 3 \quad 3 \quad 1 \\ n=4: \quad 1 \quad 4 \quad 6 \quad 4 \quad 1 \end{array} \]

8.5 Applications of the Binomial Theorem

Approximations: write a number as \((a \pm b)^n\) with \(b\) small, then expand and keep a few terms.

Remainders: to find the remainder when \(m^n\) is divided by \(d\), write \(m = d \pm k\) and expand \((d \pm k)^n\); every term except possibly the last is a multiple of \(d\).

Divisibility: the same trick shows expressions like \(5^n - 1\) are multiples of 4, since \((4+1)^n - 1\) has every remaining term divisible by 4.

Key point: In remainder and divisibility questions, always split the base as (multiple of the divisor) \(\pm\) (small number) — the expansion then does the work for you.

Mathematics 11 (Punjab Board)

Solved Examples — Chapter 8: Mathematical Induction and Binomial Theorem

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 8, worked out in full with the method explained at each step.

Example 5: Prove by mathematical induction that \(4^n + 6n - 1\) is divisible by 9 for every positive integer \(n\).

Yahan divisibility prove karni hai. Base case \(n=1\) check karenge, phir \(n=k\) assume karke \(n=k+1\) ke liye expression ko 9 ka multiple banayenge. Induction hypothesis se \(4^k\) ko replace karenge.

Base case: For \(n = 1\): \(4^1 + 6(1) - 1 = 9\), which is divisible by 9. True.

Induction hypothesis: Assume \(4^k + 6k - 1 = 9m\) for some integer \(m\), so \(4^k = 9m - 6k + 1\).

\[ 4^{k+1} + 6(k+1) - 1 = 4\cdot 4^k + 6k + 5 \]
\[ = 4(9m - 6k + 1) + 6k + 5 = 36m - 24k + 4 + 6k + 5 = 36m - 18k + 9 = 9(4m - 2k + 1) \]

This is a multiple of 9, so the statement holds for \(n = k+1\).

Answer: By the principle of mathematical induction, \(4^n + 6n - 1\) is divisible by 9 for every positive integer \(n\).

Example 6: Prove by mathematical induction that \(\sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)} = \frac{n}{2n+1}\) for every positive integer \(n\).

Series ka sum prove karna hai. Base case \(n=1\) se start karenge. Inductive step mein dono taraf \((k+1)\)-th term \(\frac{1}{(2k+1)(2k+3)}\) add karke simplify karenge.

Base case: For \(n = 1\): LHS \(= \frac{1}{1\cdot 3} = \frac{1}{3}\); RHS \(= \frac{1}{2(1)+1} = \frac{1}{3}\). True.

Induction hypothesis: Assume the formula holds for \(n = k\). Adding the \((k+1)\)-th term \(\frac{1}{(2k+1)(2k+3)}\) to both sides:

\[ \frac{k}{2k+1} + \frac{1}{(2k+1)(2k+3)} = \frac{k(2k+3) + 1}{(2k+1)(2k+3)} = \frac{2k^2+3k+1}{(2k+1)(2k+3)} \]
\[ = \frac{(2k+1)(k+1)}{(2k+1)(2k+3)} = \frac{k+1}{2k+3} = \frac{k+1}{2(k+1)+1} \]

This is exactly the formula with \(n = k+1\).

Answer: By mathematical induction, \(\sum_{k=1}^{n}\frac{1}{(2k-1)(2k+1)} = \frac{n}{2n+1}\) for every positive integer \(n\).

Example 7: Prove by the extended principle of mathematical induction that \(1 + 3 + 5 + \dots + (2n+5) = (n+3)^2\) for every integer \(n \ge -2\).

Yahan claim \(n=-2\) se start hota hai, is liye base case \(n=-2\) hoga. Baqi steps wese hi hain: \(n=k\) assume karke \(n=k+1\) prove karenge, next odd term \((2k+7)\) add karke.

Base case: For \(n = -2\): LHS \(= 2(-2)+5 = 1\); RHS \(= (-2+3)^2 = 1\). True.

Induction hypothesis: Assume \(1 + 3 + 5 + \dots + (2k+5) = (k+3)^2\) for \(k \ge -2\). Adding the next term \((2(k+1)+5) = (2k+7)\) to both sides:

\[ (k+3)^2 + (2k+7) = k^2 + 6k + 9 + 2k + 7 = k^2 + 8k + 16 = (k+4)^2 \]

Since \((k+1)+3 = k+4\), this is the formula for \(n = k+1\).

Answer: By the extended principle of mathematical induction, the formula holds for every integer \(n \ge -2\).

Example 8: Prove by mathematical induction that \(4^n \gt 3^n + 4\) for every integer \(n \ge 2\).

Inequality ka induction thora mukhtalif hai: dono taraf 4 se multiply karke phir extra positive terms ko drop kar denge. Base case \(n=2\): \(16 \gt 13\) true hai.

Base case: For \(n = 2\): \(4^2 = 16 \gt 3^2 + 4 = 13\). True.

Induction hypothesis: Assume \(4^k \gt 3^k + 4\) for \(k \ge 2\). Multiplying both sides by 4 (a positive number, so the inequality direction stays):

\[ 4^{k+1} \gt 4(3^k + 4) = 4\cdot 3^k + 16 = 3^{k+1} + 3^k + 16 \]

Since \(3^k + 16 \gt 4\) for every \(k \ge 2\), dropping the extra positive part gives \(4^{k+1} \gt 3^{k+1} + 4\).

Answer: By mathematical induction, \(4^n \gt 3^n + 4\) for every integer \(n \ge 2\).

Example 9: Prove by mathematical induction that the last digit of \(a_n = 2^{2^n} + 1\) is 7 for every integer \(n \gt 1\).

Last digit 7 ka matlab hai ke number \(10m + 7\) ki form ka hai. Base case \(n=2\): \(2^{4}+1 = 17\) ka last digit 7 hai. Inductive step mein \(2^{2^{k+1}} = (2^{2^k})^2\) likhenge aur square karenge.

Base case: For \(n = 2\): \(a_2 = 2^{2^2} + 1 = 2^4 + 1 = 17\), whose last digit is 7. True.

Induction hypothesis: Assume \(2^{2^k} + 1 = 10m + 7\) for some integer \(m\), so \(2^{2^k} = 10m + 6\). Now:

\[ 2^{2^{k+1}} + 1 = \left(2^{2^k}\right)^2 + 1 = (10m+6)^2 + 1 = 100m^2 + 120m + 36 + 1 = 10(10m^2+12m+3) + 7 \]

This ends in the digit 7.

Answer: By mathematical induction, the last digit of \(2^{2^n} + 1\) is 7 for every integer \(n \gt 1\).

Example 10: Faris deposits Rs. 1000 every month. Prove by mathematical induction that his total savings after \(n\) months are \(1000n\) rupees.

Har mahine fixed 1000 jama hota hai. Base case \(n=1\) obvious hai. Inductive step mein \((k+1)\)-th mahine ke 1000 add karke dekhenge ke formula barqarar rehta hai.

Base case: After 1 month he has deposited Rs. 1000 \(= 1000(1)\). True.

Induction hypothesis: Assume the total after \(k\) months is \(1000k\). In the \((k+1)\)-th month he deposits another Rs. 1000, so the new total is:

\[ 1000k + 1000 = 1000(k+1) \]
Answer: By mathematical induction, the total savings after \(n\) months are Rs. \(1000n\).

Example 11: Ali does 10 push-ups on the first day and increases by 2 every day. Prove by mathematical induction that his total push-ups after \(n\) days are \(n^2 + 9n\).

Day \(k\) par push-ups \(= 10 + 2(k-1) = 2k + 8\). Base case \(n=1\): \(1 + 9 = 10\) ✓. Inductive step mein \((k+1)\)-th din ke \(2(k+1)+8\) push-ups add karenge.

Base case: For \(n = 1\): total \(= 10 = 1^2 + 9(1)\). True.

Induction hypothesis: Assume the total after \(k\) days is \(k^2 + 9k\). On day \(k+1\) he does \(10 + 2k\) push-ups. New total:

\[ k^2 + 9k + (10 + 2k) = k^2 + 11k + 10 = (k+1)^2 + 9(k+1) \]

Since \((k+1)^2 + 9(k+1) = k^2 + 2k + 1 + 9k + 9 = k^2 + 11k + 10\). ✓

Answer: By mathematical induction, the total push-ups after \(n\) days are \(n^2 + 9n\).

Example 12: A person reduces daily calorie intake by 50 every week from an initial 2500. Prove by mathematical induction that the daily intake after \(n\) weeks is \(2500 - 50n\).

Har hafte 50 calories kam. Base case \(n=1\): \(2500 - 50 = 2450\) ✓. Inductive step mein agle hafte mazeed 50 subtract karenge.

Base case: After 1 week the intake is \(2500 - 50 = 2450 = 2500 - 50(1)\). True.

Induction hypothesis: Assume the intake after \(k\) weeks is \(2500 - 50k\). One more week reduces it by 50:

\[ (2500 - 50k) - 50 = 2500 - 50(k+1) \]
Answer: By mathematical induction, the daily intake after \(n\) weeks is \(2500 - 50n\) calories.

Example 13: Expand \(\left(\frac{a}{2} - \frac{2}{a}\right)^6\) using the binomial theorem and write down its general term.

Binomial theorem \((x+y)^n\) mein \(x = a/2\), \(y = -2/a\) rakhenge. General term ka formula \(T_{r+1} = \binom{n}{r}x^{n-r}y^r\) hai. Signs ka khayal rakhen kyunke \(y\) negative hai.

The general term is:

\[ T_{r+1} = \binom{6}{r}\left(\frac{a}{2}\right)^{6-r}\left(-\frac{2}{a}\right)^r = \binom{6}{r}\frac{(-1)^r 2^{2r-6}}{a^{2r-6}} \]

Writing out all seven terms (\(r = 0\) to \(6\)) and simplifying:

\[ = \frac{a^6}{64} - \frac{3a^4}{8} + \frac{15a^2}{4} - 20 + \frac{60}{a^2} - \frac{96}{a^4} + \frac{64}{a^6} \]
Answer: General term \(T_{r+1} = \binom{6}{r}\left(\frac{a}{2}\right)^{6-r}\left(-\frac{2}{a}\right)^r\); expansion \(= \frac{a^6}{64} - \frac{3a^4}{8} + \frac{15a^2}{4} - 20 + \frac{60}{a^2} - \frac{96}{a^4} + \frac{64}{a^6}\).

Example 14: Evaluate \((9.9)^5\) using the binomial theorem.

\((9.9)^5 = (10 - 0.1)^5\) likhenge. Yahan \(a = 10\), \(b = -0.1\), \(n = 5\). Expansion ke baad decimal arithmetic carefully karenge.

\[ (9.9)^5 = (10-0.1)^5 = \binom{5}{0}10^5 - \binom{5}{1}10^4(0.1) + \binom{5}{2}10^3(0.1)^2 - \binom{5}{3}10^2(0.1)^3 + \binom{5}{4}10(0.1)^4 - \binom{5}{5}(0.1)^5 \]
\[ = 100000 - 5000 + 100 - 1 + 0.005 - 0.00001 = 95099.00499 \]
Answer: \((9.9)^5 = 95099.00499\).

Example 15: In the expansion of \(\left(\frac{3}{2}x - \frac{1}{3x}\right)^{11}\), find (i) the term involving \(x^5\), (ii) the 5th term.

General term likhenge: \(T_{r+1} = \binom{11}{r}\left(\frac{3}{2}x\right)^{11-r}\left(-\frac{1}{3x}\right)^r\). (i) ke liye \(x\) ki power 5 ke barabar rakh kar \(r\) nikalenge. (ii) ke liye seedha \(r = 4\) rakhenge.

\[ T_{r+1} = (-1)^r\binom{11}{r}\frac{3^{11-2r}}{2^{11-r}}\,x^{11-2r} \]

(i) For the term in \(x^5\): \(11 - 2r = 5 \Rightarrow r = 3\).

\[ T_4 = (-1)^3\binom{11}{3}\frac{3^{5}}{2^{8}}x^5 = -\frac{165\cdot 243}{256}x^5 = -\frac{40095}{256}x^5 \]

(ii) The 5th term means \(r = 4\):

\[ T_5 = (-1)^4\binom{11}{4}\frac{3^{3}}{2^{7}}x^3 = \frac{330\cdot 27}{128}x^3 = \frac{4455}{64}x^3 \]
Answer: (i) \(-\frac{40095}{256}x^5\); (ii) \(\frac{4455}{64}x^3\).

Example 16: In the expansion of \(\left(\frac{x}{2} + \frac{2}{x^2}\right)^{12}\), find the term independent of \(x\) and the middle term.

General term mein \(x\) ki power \(12 - 3r\) hai. Independent term ke liye \(12-3r = 0\) rakhenge. \(n = 12\) even hai, to middle term \(T_{7}\) hoga (\(r = 6\)).

\[ T_{r+1} = \binom{12}{r}\left(\frac{x}{2}\right)^{12-r}\left(\frac{2}{x^2}\right)^r = \binom{12}{r}2^{2r-12}x^{12-3r} \]

Term independent of \(x\): \(12 - 3r = 0 \Rightarrow r = 4\).

\[ T_5 = \binom{12}{4}2^{-4} = \frac{495}{16} \]

Middle term (\(n = 12\) even, so \(T_{6+1} = T_7\), \(r = 6\)):

\[ T_7 = \binom{12}{6}2^{0}x^{-6} = \frac{924}{x^6} \]
Answer: Term independent of \(x\) is \(\frac{495}{16}\); middle term is \(\frac{924}{x^6}\).

Example 17: Show that \(\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \dots + n\binom{n}{n} = n\cdot 2^{n-1}\).

Yahan identity \(r\binom{n}{r} = n\binom{n-1}{r-1}\) use karenge. Har term se \(n\) common nikalne ke baad baqi sum \((1+1)^{n-1}\) ka expansion ban jayega.

Using the identity \(r\binom{n}{r} = n\binom{n-1}{r-1}\) (since \(\frac{n!}{(r-1)!(n-r)!} = n\frac{(n-1)!}{(r-1)!(n-r)!}\)):

\[ \sum_{r=1}^{n} r\binom{n}{r} = \sum_{r=1}^{n} n\binom{n-1}{r-1} = n\sum_{r=1}^{n}\binom{n-1}{r-1} \]
\[ = n\left[\binom{n-1}{0} + \binom{n-1}{1} + \dots + \binom{n-1}{n-1}\right] = n(1+1)^{n-1} = n\cdot 2^{n-1} \]
Answer: \(\binom{n}{1} + 2\binom{n}{2} + \dots + n\binom{n}{n} = n\cdot 2^{n-1}\) (proved).

Example 18: Find the general term in the expansion of \((1+x)^{-3}\) for \(|x| \lt 1\).

Negative index ka general term \(T_{r+1} = \frac{n(n-1)\dots(n-r+1)}{r!}x^r\) hai. Yahan \(n = -3\) rakhenge aur \((-1)^r\) factor nikal kar simplify karenge.

\[ T_{r+1} = \frac{(-3)(-4)(-5)\dots(-3-r+1)}{r!}x^r = \frac{(-1)^r\cdot 3\cdot 4\cdot 5\cdots(r+2)}{r!}x^r \]
\[ = (-1)^r\frac{(r+2)!}{2!\,r!}x^r = (-1)^r\binom{r+2}{2}x^r = (-1)^r\frac{(r+1)(r+2)}{2}x^r \]
Answer: \(T_{r+1} = (-1)^r\binom{r+2}{2}x^r = (-1)^r\frac{(r+1)(r+2)}{2}x^r\).

Example 19: Find the coefficient of \(x^n\) in the expansion of \(\frac{1-x}{(1+x)^2}\).

\((1-x)(1+x)^{-2}\) likhenge. \((1+x)^{-2} = \sum_{r\ge 0}(-1)^r(r+1)x^r\) hai. \(x^n\) ka coefficient do hisson se aayega: \(1\) se multiply hone wala aur \(-x\) se multiply hone wala.

\[ \frac{1-x}{(1+x)^2} = (1-x)\sum_{r\ge 0}(-1)^r(r+1)x^r \]

Coefficient of \(x^n = (-1)^n(n+1) - (-1)^{\,n-1}n = (-1)^n(n+1) + (-1)^n n = (-1)^n(2n+1)\).

Answer: \((-1)^n(2n+1)\).

Example 20: If \(x\) is so small that its cube and higher powers can be neglected, show that \(\sqrt{\frac{1-x}{1+x}} \approx 1 - x + \frac{1}{2}x^2\).

\((1-x)^{1/2}(1+x)^{-1/2}\) likhenge. Dono ko \(x^2\) tak expand karenge: \((1-x)^{1/2} \approx 1 - \frac{x}{2} - \frac{x^2}{8}\) aur \((1+x)^{-1/2} \approx 1 - \frac{x}{2} + \frac{3x^2}{8}\). Phir multiply karke \(x^3\) wale terms gira denge.

\[ \left(1 - \frac{x}{2} - \frac{x^2}{8}\right)\left(1 - \frac{x}{2} + \frac{3x^2}{8}\right) \approx 1 - x + \left(\frac{3}{8} - \frac{1}{8} + \frac{1}{4}\right)x^2 = 1 - x + \frac{1}{2}x^2 \]
Answer: \(\sqrt{\frac{1-x}{1+x}} \approx 1 - x + \frac{1}{2}x^2\) (proved).

Example 21: If \(y = \frac{1}{2}\left(\frac{4}{9}\right) + \frac{1\cdot 3}{4\cdot 2!}\left(\frac{4}{9}\right)^2 + \frac{1\cdot 3\cdot 5}{8\cdot 3!}\left(\frac{4}{9}\right)^3 + \dots\), prove that \(5y^2 + 10y - 4 = 0\).

Series ko \((1-x)^{-1/2}\) ke expansion se compare karenge. Pehla term \(\frac{1}{2}\cdot\frac{4}{9}\) batata hai ke \(x = \frac{4}{9}\) hai. Sum nikal kar di gayi equation verify karenge.

The series matches \((1-x)^{-1/2} - 1\) with \(x = \frac{4}{9}\), since the \(r\)-th term of \((1-x)^{-1/2}\) is \(\frac{1\cdot 3\cdots(2r-1)}{2^r r!}x^r\). Hence:

\[ 1 + y = \left(1 - \frac{4}{9}\right)^{-\frac{1}{2}} = \left(\frac{5}{9}\right)^{-\frac{1}{2}} = \frac{3}{\sqrt{5}} \]
\[ 5(1+y)^2 = 9 \Rightarrow 5(1 + 2y + y^2) = 9 \Rightarrow 5y^2 + 10y - 4 = 0 \]
Answer: \(5y^2 + 10y - 4 = 0\) (proved).

Example 22: Expand \((x+y)^4\) using Pascal's triangle.

Pascal's triangle ki row \(n = 4\) ke coefficients \(1, 4, 6, 4, 1\) hain. Har term mein \(x\) ki power ghatti aur \(y\) ki power barhti hai, dono ka sum 4 rehta hai.

\[ (x+y)^4 = 1\cdot x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + 1\cdot y^4 \]
Answer: \((x+y)^4 = x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\).

Example 23: Expand \((x-2)^5\) using the binomial theorem and Pascal's triangle.

\((x + (-2))^5\) likhenge. Pascal's triangle ki row \(n = 5\) ke coefficients \(1, 5, 10, 10, 5, 1\) hain. \((-2)\) ki powers ke signs alternate honge.

\[ (x-2)^5 = x^5 + 5x^4(-2) + 10x^3(-2)^2 + 10x^2(-2)^3 + 5x(-2)^4 + (-2)^5 \]
\[ = x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32 \]
Answer: \((x-2)^5 = x^5 - 10x^4 + 40x^3 - 80x^2 + 80x - 32\).

Example 24: Expand \((1-2x)^{1/3}\) to four terms and use it to evaluate \((0.8)^{1/3}\) correct to three decimal places.

\((1+x)^n\) ka formula \(n = 1/3\), \(x = -2x\) ke saath lagayenge. Chaar terms tak expand karke phir \(1-2x = 0.8\) yani \(x = 0.1\) rakhenge.

\[ (1-2x)^{\frac{1}{3}} = 1 + \frac{1}{3}(-2x) + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}(-2x)^2 + \frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}(-2x)^3 + \dots \]
\[ \approx 1 - \frac{2x}{3} - \frac{4x^2}{9} - \frac{40x^3}{81} \]

Putting \(x = 0.1\) (so that \(1 - 2x = 0.8\)):

\[ (0.8)^{\frac{1}{3}} \approx 1 - \frac{0.2}{3} - \frac{0.04}{9} - \frac{0.04}{81} \approx 1 - 0.06667 - 0.00444 - 0.00049 \approx 0.928 \]
Answer: \((0.8)^{1/3} \approx 0.928\) (correct to three decimal places).

Example 25: Evaluate \(\sqrt[3]{30}\) correct to three decimal places.

30 ke qareeb perfect cube 27 hai. \(\sqrt[3]{30} = (27+3)^{1/3} = 3\left(1+\frac{1}{9}\right)^{1/3}\) likhenge aur \((1+x)^{1/3}\) ko chaar terms tak expand karenge.

\[ \sqrt[3]{30} = 3\left(1+\frac{1}{9}\right)^{\frac{1}{3}} \approx 3\left[1 + \frac{1}{3}\cdot\frac{1}{9} - \frac{1}{9}\cdot\frac{1}{81} + \frac{5}{81}\cdot\frac{1}{729}\right] \]
\[ \approx 3\left[1 + 0.03704 - 0.00137 + 0.00008\right] \approx 3(1.03575) \approx 3.107 \]
Answer: \(\sqrt[3]{30} \approx 3.107\) (correct to three decimal places).

Example 26: Using the binomial theorem, find the remainder when \(5^{99}\) is divided by 13.

\(5^{99} = 5\cdot 5^{98} = 5\cdot 25^{49}\) likhenge. Phir \(25 = 26 - 1\) rakh kar \((26-1)^{49}\) expand karenge. 49 odd hai, to aakhri term \(-1\) hogi aur baqi sab terms 26 (aur is liye 13) ke multiples honge.

\[ 5^{99} = 5\cdot 25^{49} = 5(26-1)^{49} = 5\left[-1 + 49\cdot 26 - \binom{49}{2}26^2 + \dots\right] \]
\[ = -5 + 13\cdot m \quad \text{(for some integer } m\text{)} = 13(m-1) + 8 \]
Answer: The remainder is \(8\).

Example 27: Using the binomial theorem, show that \(11^n - 10n\) leaves remainder 1 when divided by 100, for every positive integer \(n\).

\(11 = 1 + 10\) likhenge aur \((1+10)^n\) expand karenge. Pehle do terms \(1 + 10n\) honge; teesre term se onward har term mein \(10^2 = 100\) ka factor hoga.

\[ 11^n = (1+10)^n = 1 + n\cdot 10 + \binom{n}{2}10^2 + \dots + 10^n = 1 + 10n + 100\cdot m \]

for some integer \(m\). Therefore \(11^n - 10n = 1 + 100m\).

Answer: \(11^n - 10n = 1 + 100m\), so the remainder on division by 100 is \(1\).

Example 28: Using the binomial theorem, find the last two digits of \(11^{12}\).

Last two digits ka matlab hai 100 se divide karne par remainder. \((10+1)^{12}\) expand karenge; \(100\) se multiply hone wale terms last two digits par asar nahi dalenge.

\[ 11^{12} = (10+1)^{12} = 1 + 12\cdot 10 + \binom{12}{2}10^2 + \dots = 1 + 120 + 100\cdot m = 121 + 100m \]
Answer: The last two digits are \(21\).

Example 29: Show that \(15^{13} + 13^{15}\) is divisible by 14.

\(15 = 14+1\) aur \(13 = 14-1\) likhenge. Pehle expansion ki aakhri term \(+1\), doosre ki \(-1\) (kyunke power 15 odd hai); baqi sab terms 14 ke multiples. \(+1\) aur \(-1\) cancel ho jayenge.

\[ 15^{13} = (14+1)^{13} = 1 + 14\cdot a, \qquad 13^{15} = (14-1)^{15} = -1 + 14\cdot b \]

for some integers \(a, b\). Adding: \(15^{13} + 13^{15} = 14(a+b)\), a multiple of 14.

Answer: \(15^{13} + 13^{15} = 14(a+b)\), so it is divisible by 14.

Example 30: Which number is larger: \(51^{25}\) or \(49^{25} + 50^{25}\)?

\(51^{25} = (50+1)^{25}\) expand karenge. Pehle do terms \(50^{25} + 25\cdot 50^{24}\) honge. Dikhayenge ke extra terms ka sum \(49^{25}\) se bara hai, to \(51^{25} \gt 50^{25} + 49^{25}\).

\[ 51^{25} = (50+1)^{25} = 50^{25} + 25\cdot 50^{24} + \binom{25}{2}50^{23} + \dots \]

The extra terms beyond \(50^{25}\) are all positive, and just the next two satisfy:

\[ 25\cdot 50^{24} + 300\cdot 50^{23} = 1550\cdot 50^{23} \gt 2401\cdot 49^{23} = 49^{25} \]

since \(\frac{1550}{2401} \approx 0.6456 \gt \left(\frac{49}{50}\right)^{23} \approx 0.6283\). Hence \(51^{25} \gt 50^{25} + 49^{25}\).

Answer: \(51^{25}\) is larger.

Example 31: A bank offers 5% compound interest per year. Samrina invests Rs. 100,000 for 3 years. How much will her investment be worth?

Compound interest ka formula \(A = P(1+r)^n\) hai. Yahan \(P = 100000\), \(r = 0.05\), \(n = 3\). \((1.05)^3 = (1+0.05)^3\) ko binomial theorem se expand karenge.

\[ (1.05)^3 = (1+0.05)^3 = 1 + 3(0.05) + 3(0.05)^2 + (0.05)^3 = 1 + 0.15 + 0.0075 + 0.000125 = 1.157625 \]
\[ A = 100000 \times 1.157625 = 115762.50 \]
Answer: Rs. 115,762.50.

Mathematics 11 (Punjab Board)

MCQs: Chapter 8 – Mathematical Induction and Binomial Theorem

22 multiple-choice questions covering induction steps, the binomial theorem, general and middle terms, series expansions, and applications.

  1. The first step in a proof by mathematical induction is the:

    1. Inductive step
    2. Base case
    3. Conclusion
    4. Induction hypothesis
    Base case mein \(S(1)\) (ya pehla relevant \(n\)) verify karte hain; is ke baghair induction shuru nahi hota.

  2. The number of terms in the expansion of \((a+b)^n\) (\(n\) a positive integer) is:

    1. \(n\)
    2. \(n-1\)
    3. \(n+1\)
    4. \(2n\)
    \(r = 0\) se \(r = n\) tak \(n+1\) terms hote hain.

  3. The general term \(T_{r+1}\) in the expansion of \((a+b)^n\) is:

    1. \(\binom{n}{r}a^{n-r}b^r\)
    2. \(\binom{n}{r}a^rb^{n-r}\)
    3. \(\binom{n}{r+1}a^{n-r}b^r\)
    4. \(\binom{n}{r}a^nb^r\)
    \(T_{r+1} = \binom{n}{r}a^{n-r}b^r\) — \(a\) ki power ghatti aur \(b\) ki barhti hai, sum \(n\) rehta hai.

  4. The middle term in the expansion of \((x+y)^8\) is:

    1. \(56x^4y^4\)
    2. \(70x^4y^4\)
    3. \(70x^5y^3\)
    4. \(28x^4y^4\)
    \(n = 8\) even hai, to middle term \(T_5 = \binom{8}{4}x^4y^4 = 70x^4y^4\).

  5. The binomial series for \((1+x)^n\) (negative or fractional \(n\)) is valid when:

    1. \(|x| \gt 1\)
    2. \(x = 1\)
    3. \(|x| \lt 1\)
    4. \(x \ge 0\)
    Infinite binomial series sirf \(|x| \lt 1\) ke liye converge karti hai.

  6. The coefficient of \(x^2\) in the expansion of \((1+x)^{-1}\) is:

    1. \(1\)
    2. \(-1\)
    3. \(2\)
    4. \(-2\)
    \((1+x)^{-1} = 1 - x + x^2 - x^3 + \dots\), to \(x^2\) ka coefficient \(1\) hai.

  7. \((1-x)^{-1}\) equals:

    1. \(1-x+x^2-x^3+\dots\)
    2. \(1-2x+3x^2-4x^3+\dots\)
    3. \(1+x-x^2+x^3-\dots\)
    4. \(1+x+x^2+x^3+\dots\)
    \((1-x)^{-1}\) geometric series hai jiska common ratio \(x\) hai: \(1 + x + x^2 + \dots\).

  8. If \(n\) is odd, the number of terms in \((a+b)^n + (a-b)^n\) is:

    1. \(n+1\)
    2. \(\frac{n+1}{2}\)
    3. \(n\)
    4. \(\frac{n-1}{2}\)
    Odd powers cancel ho jati hain; \(n\) odd hone par \(\frac{n+1}{2}\) even-powered terms bachte hain.

  9. \(5^5 + 7^5\) is divisible by:

    1. \(36\)
    2. \(18\)
    3. \(12\)
    4. \(7\)
    \((6-1)^5 + (6+1)^5 = 12\times 1661\); ye 12 se divisible hai (36 se nahi).

  10. The remainder when \(8^{100}\) is divided by 7 is:

    1. \(1\)
    2. \(7\)
    3. \(0\)
    4. \(2\)
    \(8^{100} = (7+1)^{100} = 7m + 1\), to remainder \(1\) hai.

  11. \(\binom{n}{0} + \binom{n}{1} + \dots + \binom{n}{n}\) equals:

    1. \(n^2\)
    2. \(2n\)
    3. \(n!\)
    4. \(2^n\)
    \((1+1)^n = 2^n\) mein \(x = 1\) rakhne se ye sum milta hai.

  12. \(\binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}\) equals:

    1. \(2^n\)
    2. \(2^n - 1\)
    3. \(2^{n-1}\)
    4. \(n\cdot 2^{n-1}\)
    Total sum \(2^n\) mein se \(\binom{n}{0} = 1\) subtract kar den: \(2^n - 1\).

  13. The term independent of \(x\) in \(\left(x+\frac{1}{x}\right)^6\) is:

    1. \(15\)
    2. \(6\)
    3. \(20\)
    4. \(1\)
    \(T_{r+1} = \binom{6}{r}x^{6-2r}\); \(6-2r = 0\) se \(r = 3\), aur \(\binom{6}{3} = 20\).

  14. Using the binomial theorem, \((0.99)^2\) is approximately:

    1. \(0.9801\)
    2. \(0.981\)
    3. \(0.99\)
    4. \(0.98\)
    \((1-0.01)^2 = 1 - 0.02 + 0.0001 = 0.9801\).

  15. \(n! \gt 2^{n-1}\) holds for:

    1. \(n \ge 1\)
    2. \(n \ge 2\)
    3. \(n \ge 3\)
    4. \(n \ge 4\)
    \(n = 3\) par \(6 \lt 8\); pehli baar \(n = 4\) par \(24 \gt 8\) true hota hai.

  16. The coefficients in row \(n = 4\) of Pascal's triangle are:

    1. \(1, 3, 3, 1\)
    2. \(1, 4, 6, 4, 1\)
    3. \(1, 5, 10, 10, 5, 1\)
    4. \(1, 2, 1\)
    Row 4: \(\binom{4}{0}, \binom{4}{1}, \binom{4}{2}, \binom{4}{3}, \binom{4}{4} = 1, 4, 6, 4, 1\).

  17. For small \(x\), \((1+x)^{1/2}\) is approximately:

    1. \(1+\frac{x}{2}\)
    2. \(1-\frac{x}{2}\)
    3. \(1+2x\)
    4. \(1+x\)
    \((1+x)^n \approx 1+nx\) mein \(n = 1/2\) rakhen: \(1 + x/2\).

  18. The unit digit of \(2^{32}\) is:

    1. \(2\)
    2. \(4\)
    3. \(6\)
    4. \(8\)
    2 ki powers ke unit digits 2, 4, 8, 6 repeat hote hain; \(32 \div 4\) ka remainder 0 hai, to digit \(6\).

  19. The coefficient of \(x^r\) equals the coefficient of \(x^{n-r}\) in \((1+x)^n\) because of:

    1. Symmetry \(\binom{n}{r} = \binom{n}{n-r}\)
    2. Pascal's identity
    3. Mathematical induction
    4. The remainder theorem
    Binomial coefficients symmetric hote hain: \(\binom{n}{r} = \binom{n}{n-r}\).

  20. \((2+\sqrt{3})^2 + (2-\sqrt{3})^2\) equals:

    1. \(7\)
    2. \(14\)
    3. \(28\)
    4. \(4\)
    \((4+4\sqrt{3}+3) + (4-4\sqrt{3}+3) = 14\); \(\sqrt{3}\) wale terms cancel.

  21. The term independent of \(x\) in \(\left(x^2-\frac{1}{x}\right)^6\) is:

    1. \(6\)
    2. \(20\)
    3. \(-15\)
    4. \(15\)
    \(T_{r+1} = \binom{6}{r}(-1)^r x^{12-3r}\); \(12-3r = 0\) se \(r = 4\): \(\binom{6}{4} = 15\).

  22. Using the binomial theorem, \((1.02)^3\) is approximately:

    1. \(1.06\)
    2. \(1.061\)
    3. \(1.0612\)
    4. \(1.062\)
    \((1+0.02)^3 = 1 + 0.06 + 3(0.0004) + 0.000008 = 1.061208 \approx 1.0612\).

Mathematics 11 (Punjab Board)

Chapter 8: Mathematical Induction and Binomial Theorem — Formula Sheet

All the key results of Chapter 8 in one place — induction steps, binomial expansions, general terms, and the standard series.

Mathematical Induction

\[ \text{Step 1 (base case): prove } S(n_0). \quad \text{Step 2: } S(k) \Rightarrow S(k+1). \]
\[ \text{Extended induction: base case at the first integer where the claim starts.} \]

Binomial Theorem (Positive Integral Index)

\[ (a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}, \qquad n \in \mathbb{N} \]
\[ \binom{n}{r} = \frac{n!}{r!(n-r)!}, \qquad \text{number of terms} = n+1 \]
\[ T_{r+1} = \binom{n}{r}a^{n-r}b^{r} \quad \text{(general term)} \]

Middle Terms

\[ n \text{ even: middle term } = T_{\frac{n}{2}+1} \]
\[ n \text{ odd: middle terms } = T_{\frac{n+1}{2}} \text{ and } T_{\frac{n+3}{2}} \]

Binomial Coefficient Identities

\[ \binom{n}{r} = \binom{n}{n-r} \]
\[ \binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r} \quad \text{(Pascal's identity)} \]
\[ r\binom{n}{r} = n\binom{n-1}{r-1} \]
\[ \sum_{r=0}^{n}\binom{n}{r} = 2^n, \qquad \sum_{r=1}^{n}\binom{n}{r} = 2^n - 1 \]
\[ \sum_{r=1}^{n}r\binom{n}{r} = n\cdot 2^{n-1} \]
\[ (a+b)^n + (a-b)^n = 2\left[\binom{n}{0}a^n + \binom{n}{2}a^{n-2}b^2 + \dots\right] \]

Binomial Series (Negative or Fractional Index, \(|x| \lt 1\))

\[ (1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots \]
\[ T_{r+1} = \frac{n(n-1)\cdots(n-r+1)}{r!}x^r \]
\[ (1+x)^{-1} = 1 - x + x^2 - x^3 + \dots \]
\[ (1-x)^{-1} = 1 + x + x^2 + x^3 + \dots \]
\[ (1+x)^{-2} = 1 - 2x + 3x^2 - 4x^3 + \dots \]
\[ (1-x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots \]
\[ (1+x)^{\frac{1}{2}} = 1 + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{16} - \dots \]
\[ (1+x)^{-\frac{1}{2}} = 1 - \frac{x}{2} + \frac{3x^2}{8} - \frac{5x^3}{16} + \dots \]

Approximations (Small \(x\))

\[ (1+x)^n \approx 1 + nx \quad \text{(neglecting } x^2 \text{ and higher)} \]
\[ (a+b)^n = a^n\left(1+\frac{b}{a}\right)^n, \quad \left|\frac{b}{a}\right| \lt 1 \]

Remainder and Divisibility Tricks

\[ (d\pm k)^n = (\text{multiple of } d) \pm k^n \]
\[ 11^n - 10n = 1 + 100m, \qquad 5^n - 1 = 4m, \qquad 7^n - 6n = 1 + 6m \]

Chapter 8 FAQs

What is covered in Chapter 8 (Mathematical Induction and Binomial Theorem)?
This chapter covers Principle of mathematical induction, binomial expansion, general and middle term. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.
How many marks does Chapter 8 carry in the Punjab Board paper?
Chapter 8 (Mathematical Induction and Binomial Theorem) carries approximately 11 marks in the Punjab Board Class 11 Mathematics paper.
Are these notes for the new PCTB syllabus?
Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.
Where are the exercise solutions for Chapter 8?
Use the exercise index on this page — every exercise of Chapter 8 has its own page with complete step-by-step solutions.