Mathematics 11 (Punjab Board)
Exercise 8.1 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 8: Mathematical Induction and Binomial Theorem. Every proof follows the same pattern — base case, induction hypothesis, then the \(k \to k+1\) step.
Question 1(i): Use mathematical induction to prove that \(\log x^n = n\log x\) for every positive integer \(n\), where \(x\) is positive.
Logarithm ka power rule induction se prove karenge. Base case \(n=1\) obvious hai. Inductive step mein \(x^{k+1} = x^k\cdot x\) likh kar \(\log(ab) = \log a + \log b\) use karenge.
Base case: For \(n = 1\): \(\log x^1 = \log x = 1\cdot\log x\). True.
Induction hypothesis: Assume \(\log x^k = k\log x\) for some positive integer \(k\).
For \(n = k+1\): using the product rule \(\log(ab) = \log a + \log b\):
\[
\log x^{k+1} = \log(x^k\cdot x) = \log x^k + \log x = k\log x + \log x = (k+1)\log x
\]
Answer: By mathematical induction, \(\log x^n = n\log x\) for every positive integer \(n\).
Question 1(ii): Use mathematical induction to prove that \(2 + 5 + 8 + \dots + (3n-1) = \frac{n}{2}(3n+1)\) for every positive integer \(n\).
Ye arithmetic series hai jiska general term \((3n-1)\) hai. Base case \(n=1\) check karenge. Inductive step mein dono taraf next term \((3k+2)\) add karke factorize karenge.
Base case: For \(n = 1\): LHS \(= 2\); RHS \(= \frac{1}{2}(3+1) = 2\). True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding the \((k+1)\)-th term, which is \(3(k+1)-1 = 3k+2\), to both sides:
\[
\frac{k}{2}(3k+1) + (3k+2) = \frac{3k^2+k+6k+4}{2} = \frac{3k^2+7k+4}{2} = \frac{(k+1)(3k+4)}{2}
\]
Since \(3k+4 = 3(k+1)+1\), this equals \(\frac{k+1}{2}(3(k+1)+1)\), the formula for \(n = k+1\).
Answer: By mathematical induction, \(2 + 5 + 8 + \dots + (3n-1) = \frac{n}{2}(3n+1)\) for every positive integer \(n\).
Question 1(iii): Use mathematical induction to prove that \(2 + (2+5) + (2+5+8) + \dots + \frac{n}{2}(3n+1) = \frac{n}{4}(n+1)^2\) for every positive integer \(n\).
Note: the printed right–hand side \(\frac{n}{4}(n+1)^2\) fails the base case (\(n=1\) gives \(2 \ne 1\)). The identity that is actually true — and is proved below — is \(2 + (2+5) + \dots + \frac{n}{2}(3n+1) = \frac{n}{2}(n+1)^2\).
Base case: For \(n = 1\): LHS \(= 2\); RHS \(= \frac{1}{2}(2)^2 = 2\). True.
Induction hypothesis: Assume the (corrected) formula holds for \(n = k\). Adding the \((k+1)\)-th term \(\frac{k+1}{2}(3k+4)\) to both sides:
\[
\frac{k}{2}(k+1)^2 + \frac{k+1}{2}(3k+4) = \frac{k+1}{2}\left[k(k+1) + (3k+4)\right] = \frac{k+1}{2}\left[k^2+4k+4\right]
\]
\[
= \frac{k+1}{2}(k+2)^2
\]
This is the formula with \(n = k+1\).
Answer: By mathematical induction, \(2 + (2+5) + (2+5+8) + \dots + \frac{n}{2}(3n+1) = \frac{n}{2}(n+1)^2\) for every positive integer \(n\).
Question 1(iv): Use mathematical induction to prove that \(2 + 6 + 18 + \dots + 2\times 3^{n-1} = 3^n - 1\) for every positive integer \(n\).
Ye geometric series hai jiska common ratio 3 hai. Base case \(n=1\): \(2 = 3-1\) ✓. Inductive step mein next term \(2\cdot 3^k\) add karenge.
Base case: For \(n = 1\): LHS \(= 2\); RHS \(= 3^1 - 1 = 2\). True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding the \((k+1)\)-th term \(2\cdot 3^k\) to both sides:
\[
(3^k - 1) + 2\cdot 3^k = 3\cdot 3^k - 1 = 3^{k+1} - 1
\]
Answer: By mathematical induction, \(2 + 6 + 18 + \dots + 2\times 3^{n-1} = 3^n - 1\) for every positive integer \(n\).
Question 1(v): Use mathematical induction to prove that \(1\times 3 + 2\times 5 + 3\times 7 + \dots + n(2n+1) = \frac{n(n+1)(4n+5)}{6}\) for every positive integer \(n\).
General term \(n(2n+1)\) hai. Base case \(n=1\): \(3 = \frac{1\cdot 2\cdot 9}{6}\) ✓. Inductive step mein \((k+1)(2k+3)\) add karke \((k+2)(4k+9)\) factor nikalenge.
Base case: For \(n = 1\): LHS \(= 3\); RHS \(= \frac{1\cdot 2\cdot 9}{6} = 3\). True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding the \((k+1)\)-th term \((k+1)(2k+3)\):
\[
\frac{k(k+1)(4k+5)}{6} + (k+1)(2k+3) = \frac{k+1}{6}\left[k(4k+5) + 6(2k+3)\right]
\]
\[
= \frac{k+1}{6}\left[4k^2+17k+18\right] = \frac{k+1}{6}(k+2)(4k+9) = \frac{(k+1)(k+2)(4(k+1)+5)}{6}
\]
Answer: By mathematical induction, the formula holds for every positive integer \(n\).
Question 1(vi): Use mathematical induction to prove that \(\frac{1}{1\times 2} + \frac{1}{2\times 3} + \dots + \frac{1}{n(n+1)} = 1 - \frac{1}{n+1}\) for every positive integer \(n\).
Telescoping jaisi series hai. Base case \(n=1\): \(\frac{1}{2} = 1-\frac{1}{2}\) ✓. Inductive step mein \(\frac{1}{(k+1)(k+2)}\) add karke common denominator lenge.
Base case: For \(n = 1\): LHS \(= \frac{1}{2}\); RHS \(= 1 - \frac{1}{2} = \frac{1}{2}\). True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding \(\frac{1}{(k+1)(k+2)}\) to both sides:
\[
\left(1 - \frac{1}{k+1}\right) + \frac{1}{(k+1)(k+2)} = 1 - \frac{(k+2)-1}{(k+1)(k+2)} = 1 - \frac{k+1}{(k+1)(k+2)} = 1 - \frac{1}{k+2}
\]
Answer: By mathematical induction, the formula holds for every positive integer \(n\).
Question 1(vii): Use mathematical induction to prove that \(r + r^2 + r^3 + \dots + r^n = \frac{r(1-r^n)}{1-r}\) for every positive integer \(n\), where \(r \ne 1\).
Ye geometric series ka sum formula hai. Base case \(n=1\) direct hai. Inductive step mein \(r^{k+1}\) add karke numerator mein se \(r^{k+1}\) common nikalenge.
Base case: For \(n = 1\): RHS \(= \frac{r(1-r)}{1-r} = r =\) LHS. True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding \(r^{k+1}\) to both sides:
\[
\frac{r(1-r^k)}{1-r} + r^{k+1} = \frac{r - r^{k+1} + r^{k+1}(1-r)}{1-r} = \frac{r - r^{k+2}}{1-r} = \frac{r(1-r^{k+1})}{1-r}
\]
Answer: By mathematical induction, the geometric sum formula holds for every positive integer \(n\) (with \(r \ne 1\)).
Question 1(viii): Use mathematical induction to prove that \(a + (a+d) + (a+2d) + \dots + [a+(n-1)d] = \frac{n}{2}[2a+(n-1)d]\) for every positive integer \(n\).
Arithmetic series ka sum formula. Base case \(n=1\) dono taraf \(a\) deta hai. Inductive step mein next term \(a+kd\) add karke \((k+1)\) common nikalenge.
Base case: For \(n = 1\): LHS \(= a\); RHS \(= \frac{1}{2}(2a) = a\). True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding the \((k+1)\)-th term \(a + kd\):
\[
\frac{k}{2}[2a+(k-1)d] + (a+kd) = \frac{2ak + k(k-1)d + 2a + 2kd}{2} = \frac{2a(k+1) + kd(k+1)}{2}
\]
\[
= \frac{k+1}{2}[2a+kd] = \frac{k+1}{2}[2a+((k+1)-1)d]
\]
Answer: By mathematical induction, the A.P. sum formula holds for every positive integer \(n\).
Question 1(ix): Use mathematical induction to prove that \(a_n = a_1 + (n-1)d\) for every positive integer \(n\), when \(a_1, a_1+d, a_1+2d, \dots\) form an A.P.
A.P. mein har term pichle term mein \(d\) add karke milta hai, yani \(a_{k+1} = a_k + d\). Is recurrence ko induction hypothesis ke saath use karenge.
Base case: For \(n = 1\): \(a_1 = a_1 + (1-1)d = a_1\). True.
Induction hypothesis: Assume \(a_k = a_1 + (k-1)d\). Since consecutive terms of an A.P. differ by \(d\):
\[
a_{k+1} = a_k + d = [a_1 + (k-1)d] + d = a_1 + kd = a_1 + ((k+1)-1)d
\]
Answer: By mathematical induction, \(a_n = a_1 + (n-1)d\) for every positive integer \(n\).
Question 1(x): Use mathematical induction to prove that \(a_n = a_1r^{n-1}\) for every positive integer \(n\), when \(a_1, a_1r, a_1r^2, \dots\) form a G.P.
G.P. mein \(a_{k+1} = a_k\cdot r\). Induction hypothesis mein \(r\) se multiply karke powers ka rule \(r^{k-1}\cdot r = r^k\) lagayenge.
Base case: For \(n = 1\): \(a_1 = a_1r^{0} = a_1\). True.
Induction hypothesis: Assume \(a_k = a_1r^{k-1}\). Then:
\[
a_{k+1} = a_k\cdot r = (a_1r^{k-1})\cdot r = a_1r^k = a_1r^{(k+1)-1}
\]
Answer: By mathematical induction, \(a_n = a_1r^{n-1}\) for every positive integer \(n\).
Question 1(xi): Use mathematical induction to prove that \(\binom{3}{3} + \binom{4}{3} + \binom{5}{3} + \dots + \binom{n+2}{3} = \binom{n+3}{4}\) for every positive integer \(n\).
Yahan Pascal's identity \(\binom{m}{r} + \binom{m}{r-1} = \binom{m+1}{r}\) kaam aayegi. Base case \(n=1\): \(\binom{3}{3} = \binom{4}{4} = 1\) ✓.
Base case: For \(n = 1\): LHS \(= \binom{3}{3} = 1\); RHS \(= \binom{4}{4} = 1\). True.
Induction hypothesis: Assume the sum up to \(\binom{k+2}{3}\) equals \(\binom{k+3}{4}\). Adding the next term \(\binom{k+3}{3}\) and applying Pascal's identity:
\[
\binom{k+3}{4} + \binom{k+3}{3} = \binom{k+4}{4} = \binom{(k+1)+3}{4}
\]
Answer: By mathematical induction, the identity holds for every positive integer \(n\).
Question 1(xii): Use mathematical induction to prove that the sum of the first \(n\) odd natural numbers is \(n^2\).
Pehle \(n\) odd numbers \(1, 3, 5, \dots, (2n-1)\) hain. Base case \(n=1\) clear hai. Inductive step mein \((k+1)\)-th odd number \((2k+1)\) add karke perfect square banayenge.
Base case: For \(n = 1\): sum \(= 1 = 1^2\). True.
Induction hypothesis: Assume \(1 + 3 + \dots + (2k-1) = k^2\). Adding the next odd number \((2k+1)\):
\[
k^2 + (2k+1) = (k+1)^2
\]
Answer: By mathematical induction, the sum of the first \(n\) odd natural numbers is \(n^2\).
Question 2(i): Prove by mathematical induction that \(n^2 + n\) is divisible by 2 for every positive integer \(n\).
Divisibility ke liye expression ko 2 ka multiple likhna hai. \((k+1)^2 + (k+1)\) expand karke induction hypothesis \((k^2+k = 2m)\) substitute karenge.
Base case: For \(n = 1\): \(1^2 + 1 = 2\), divisible by 2. True.
Induction hypothesis: Assume \(k^2 + k = 2m\) for some integer \(m\). Then:
\[
(k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = (k^2+k) + 2k + 2 = 2m + 2(k+1) = 2(m+k+1)
\]
which is divisible by 2.
Answer: By mathematical induction, \(n^2 + n\) is divisible by 2 for every positive integer \(n\).
Question 2(ii): Prove by mathematical induction that \(5^n - 2^n\) is divisible by 3 for every positive integer \(n\).
\(5^{k+1} - 2^{k+1}\) ko aise likhenge ke \(5^k - 2^k\) ka group ban jaye: \(5\cdot 5^k - 5\cdot 2^k + 5\cdot 2^k - 2\cdot 2^k\). Pehla group hypothesis se, baqi \(3\cdot 2^k\) ban jayega.
Base case: For \(n = 1\): \(5 - 2 = 3\), divisible by 3. True.
Induction hypothesis: Assume \(5^k - 2^k = 3m\) for some integer \(m\). Then:
\[
5^{k+1} - 2^{k+1} = 5\cdot 5^k - 2\cdot 2^k = 5(5^k - 2^k) + 3\cdot 2^k = 5(3m) + 3\cdot 2^k = 3(5m + 2^k)
\]
which is divisible by 3.
Answer: By mathematical induction, \(5^n - 2^n\) is divisible by 3 for every positive integer \(n\).
Question 2(iii): Prove by mathematical induction that \(8\times 10^n - 2\) is divisible by 6 for every positive integer \(n\).
Base case \(n=1\): \(80 - 2 = 78 = 6\times 13\) ✓. Inductive step mein \(8\cdot 10^{k+1} - 2 = 10(8\cdot 10^k - 2) + 18\) likhenge; dono hisse 6 se divisible honge.
Base case: For \(n = 1\): \(8(10) - 2 = 78 = 6\times 13\). True.
Induction hypothesis: Assume \(8\cdot 10^k - 2 = 6m\) for some integer \(m\). Then:
\[
8\cdot 10^{k+1} - 2 = 10(8\cdot 10^k) - 2 = 10(8\cdot 10^k - 2) + 18 = 10(6m) + 18 = 6(10m+3)
\]
which is divisible by 6.
Answer: By mathematical induction, \(8\times 10^n - 2\) is divisible by 6 for every positive integer \(n\).
Question 3: Prove by mathematical induction that \(\sum_{k=1}^{n} r^k = \frac{r^{n+1}-1}{r-1} - 1\) for every positive integer \(n\) (with \(r \ne 1\)).
Ye geometric series \(r + r^2 + \dots + r^n\) ka doosra roop hai. Base case \(n=1\) par RHS simplify karke \(r\) milta hai. Inductive step mein \(r^{k+1}\) add karke common denominator lenge.
Base case: For \(n = 1\): RHS \(= \frac{r^2-1}{r-1} - 1 = (r+1) - 1 = r =\) LHS. True.
Induction hypothesis: Assume the formula holds for \(n = k\). Adding \(r^{k+1}\) to both sides:
\[
\frac{r^{k+1}-1}{r-1} - 1 + r^{k+1} = \frac{r^{k+1}-1 + r^{k+1}(r-1) - (r-1)}{r-1} = \frac{r^{k+2}-r}{r-1} = \frac{r^{(k+1)+1}-1}{r-1} - 1
\]
Answer: By mathematical induction, \(\sum_{k=1}^{n} r^k = \frac{r^{n+1}-1}{r-1} - 1\) for every positive integer \(n\).
Question 4: Prove by mathematical induction that \(x - y\) is a factor of \(x^n - y^n\) for every positive integer \(n\) (with \(x \ne y\)).
Factor ka matlab hai ke \(x^n - y^n = (x-y)\times\) (koi expression). \(x^{k+1} - y^{k+1}\) ko \(x\cdot x^k - x\cdot y^k + x\cdot y^k - y\cdot y^k\) likh kar dono groups mein se \((x-y)\) nikalenge.
Base case: For \(n = 1\): \(x - y\) is a factor of itself. True.
Induction hypothesis: Assume \(x^k - y^k = (x-y)m\) for some expression \(m\). Then:
\[
x^{k+1} - y^{k+1} = x\cdot x^k - y\cdot y^k = x(x^k - y^k) + y^k(x - y) = x(x-y)m + y^k(x-y) = (x-y)(xm + y^k)
\]
So \(x - y\) is a factor of \(x^{k+1} - y^{k+1}\).
Answer: By mathematical induction, \(x - y\) is a factor of \(x^n - y^n\) for every positive integer \(n\).
Question 5: Prove by mathematical induction that \(n! \gt 2^{n-1}\) for every integer \(n \ge 4\).
Extended induction hai: base case \(n=4\) (\(24 \gt 8\)). Inductive step mein \((k+1)! = (k+1)\cdot k!\) likhenge aur \(k+1 \gt 2\) use karenge (kyunke \(k \ge 4\)).
Base case: For \(n = 4\): \(4! = 24 \gt 2^3 = 8\). True.
Induction hypothesis: Assume \(k! \gt 2^{k-1}\) for \(k \ge 4\). Then:
\[
(k+1)! = (k+1)\cdot k! \gt (k+1)\cdot 2^{k-1} \ge 5\cdot 2^{k-1} \gt 2\cdot 2^{k-1} = 2^k
\]
since \(k + 1 \ge 5 \gt 2\).
Answer: By mathematical induction, \(n! \gt 2^{n-1}\) for every integer \(n \ge 4\).
Question 6: Prove by mathematical induction that \(4^n \gt 3^n + 2^{n+1}\) for integral values of \(n \ge 2\).
Note: the printed claim fails at \(n = 2\) (\(16 \lt 17\)), so the inequality actually holds for \(n \ge 3\). The induction below uses base case \(n = 3\) (\(64 \gt 43\)).
Base case: For \(n = 3\): \(4^3 = 64 \gt 3^3 + 2^4 = 27 + 16 = 43\). True.
Induction hypothesis: Assume \(4^k \gt 3^k + 2^{k+1}\) for \(k \ge 3\). Multiplying by 4:
\[
4^{k+1} = 4\cdot 4^k \gt 4(3^k + 2^{k+1}) = 4\cdot 3^k + 2^{k+3} \gt 3\cdot 3^k + 2\cdot 2^{k+1} = 3^{k+1} + 2^{k+2}
\]
since \(4\cdot 3^k \gt 3\cdot 3^k\) and \(2^{k+3} = 4\cdot 2^{k+1} \gt 2\cdot 2^{k+1}\).
Answer: By mathematical induction, \(4^n \gt 3^n + 2^{n+1}\) for every integer \(n \ge 3\).
Question 7: Prove by mathematical induction that \(1 + nx \le (1+x)^n\) for \(n \ge 2\) and \(x \gt -1\).
Ye Bernoulli's inequality hai. Base case \(n=2\): \((1+x)^2 = 1+2x+x^2 \ge 1+2x\) ✓. Inductive step mein dono taraf \((1+x)\) se multiply karenge (jo positive hai, is liye inequality ki direction nahi badlegi).
Base case: For \(n = 2\): \((1+x)^2 = 1 + 2x + x^2 \ge 1 + 2x\) since \(x^2 \ge 0\). True.
Induction hypothesis: Assume \((1+x)^k \ge 1 + kx\) for \(k \ge 2\). Since \(x \gt -1\), we have \(1 + x \gt 0\); multiplying both sides by \(1+x\):
\[
(1+x)^{k+1} \ge (1+x)(1+kx) = 1 + kx + x + kx^2 = 1 + (k+1)x + kx^2 \ge 1 + (k+1)x
\]
since \(kx^2 \ge 0\).
Answer: By mathematical induction, \(1 + nx \le (1+x)^n\) for \(n \ge 2\) and \(x \gt -1\).
Question 8: Aliza invests Rs. 1,000,000 in a business that promises a 6% return compounded annually. Prove by mathematical induction that the amount after \(n\) years is \(1{,}000{,}000(1.06)^n\).
Compound growth mein har saal amount 1.06 se multiply hota hai. Base case \(n=1\): \(1{,}000{,}000 \times 1.06\) ✓. Inductive step mein \(A(k+1) = 1.06\times A(k)\) use karenge.
Base case: After 1 year the amount is \(1{,}000{,}000(1 + 0.06) = 1{,}000{,}000(1.06)^1\). True.
Induction hypothesis: Assume the amount after \(k\) years is \(1{,}000{,}000(1.06)^k\). One more year of 6% growth multiplies it by 1.06:
\[
A(k+1) = 1.06\times\left[1{,}000{,}000(1.06)^k\right] = 1{,}000{,}000(1.06)^{k+1}
\]
Answer: By mathematical induction, the amount after \(n\) years is Rs. \(1{,}000{,}000(1.06)^n\).
Question 9: A bank offers annual interest rate \(r\). If \(P\) rupees are invested, the amount after \(n\) years is \(A(n) = P(1+r)^n\). Prove by induction that this formula holds for all \(n \ge 0\).
Yahan \(n = 0\) se start hai: \(A(0) = P(1+r)^0 = P\) (initial investment) ✓. Baqi proof Question 8 jaisa hai, sirf \(P\) aur \(r\) general hain.
Base case: For \(n = 0\): \(A(0) = P(1+r)^0 = P\), the amount initially invested. True.
Induction hypothesis: Assume \(A(k) = P(1+r)^k\). After one more year at rate \(r\):
\[
A(k+1) = (1+r)\cdot A(k) = (1+r)\cdot P(1+r)^k = P(1+r)^{k+1}
\]
Answer: By mathematical induction, \(A(n) = P(1+r)^n\) for all \(n \ge 0\).
Question 10: Sikandar saves Rs. 500 in the first month and increases his savings by Rs. 500 every subsequent month. Using mathematical induction, determine whether his total savings will reach at least Rs. 12,000 after 24 months.
Month \(k\) ki saving \(500k\) hai. Pehle total savings ka formula \(S(n) = 250n(n+1)\) induction se prove karenge, phir \(n = 24\) rakhenge.
Claim: Total savings after \(n\) months are \(S(n) = 500 + 1000 + \dots + 500n = 250n(n+1)\).
Base case: For \(n = 1\): \(S(1) = 500 = 250(1)(2)\). True.
Induction hypothesis: Assume \(S(k) = 250k(k+1)\). In month \(k+1\) he saves \(500(k+1)\):
\[
S(k+1) = 250k(k+1) + 500(k+1) = (k+1)(250k + 500) = 250(k+1)(k+2)
\]
Check for 24 months: \(S(24) = 250\times 24\times 25 = 150{,}000 \ge 12{,}000\).
Answer: Yes — his total savings after 24 months will be Rs. 150,000, which is well above Rs. 12,000.
Question 11: Prove by mathematical induction that if Ali takes a loan of Rs. 2,000,000 and pays Rs. 50,000 at the end of each year, the remaining balance after \(n\) years is \(R_n = 2{,}000{,}000 - 50{,}000n\).
Har saal ke aakhir mein balance 50,000 kam hota hai. Base case \(n=1\): \(2{,}000{,}000 - 50{,}000 = 1{,}950{,}000\) ✓. Inductive step mein \(R_{k+1} = R_k - 50{,}000\).
Base case: After 1 year: \(2{,}000{,}000 - 50{,}000 = 1{,}950{,}000 = 2{,}000{,}000 - 50{,}000(1)\). True.
Induction hypothesis: Assume \(R_k = 2{,}000{,}000 - 50{,}000k\). After one more annual payment:
\[
R_{k+1} = R_k - 50{,}000 = 2{,}000{,}000 - 50{,}000k - 50{,}000 = 2{,}000{,}000 - 50{,}000(k+1)
\]
Answer: By mathematical induction, \(R_n = 2{,}000{,}000 - 50{,}000n\) after \(n\) years.
Question 12: Salman starts saving with Rs. 5,000 and saves an additional Rs. 1,000 at the end of every month. Derive a formula \(S(n)\) for his total savings after \(n\) months, and prove it by mathematical induction.
Initial 5,000 plus har mahine 1,000: formula \(S(n) = 5{,}000 + 1{,}000n\) banta hai. Ise induction se prove karenge: base case \(n=1\) par \(6{,}000\) hona chahiye.
Formula: \(S(n) = 5{,}000 + 1{,}000n\).
Base case: After 1 month: \(5{,}000 + 1{,}000 = 6{,}000 = 5{,}000 + 1{,}000(1)\). True.
Induction hypothesis: Assume \(S(k) = 5{,}000 + 1{,}000k\). One more month adds Rs. 1,000:
\[
S(k+1) = S(k) + 1{,}000 = 5{,}000 + 1{,}000k + 1{,}000 = 5{,}000 + 1{,}000(k+1)
\]
Answer: \(S(n) = 5{,}000 + 1{,}000n\), proved by mathematical induction.
