Mathematics 11 (Punjab Board)
Exercise 8.4 Solutions
Worked solutions for Punjab Board Class 11 Mathematics, Chapter 8: Mathematical Induction and Binomial Theorem. This exercise applies the binomial theorem to approximations, remainders, divisibility, and comparisons.
Question 1: Use the binomial theorem to find an approximate value of \(\sqrt{99}\).
\(\sqrt{99} = (100-1)^{1/2} = 10\left(1-\frac{1}{100}\right)^{1/2}\) likhenge. Phir \((1+x)^{1/2}\) ko teen terms tak expand karke \(x = -1/100\) rakhenge.
\[
\sqrt{99} = 10\left(1-\frac{1}{100}\right)^{\frac{1}{2}} \approx 10\left[1 + \frac{1}{2}\left(-\frac{1}{100}\right) - \frac{1}{8}\left(\frac{1}{100}\right)^2\right]
\]
\[
= 10\left[1 - 0.005 - 0.0000125\right] = 10(0.9949875) = 9.949875 \approx 9.95
\]
Answer: \(\sqrt{99} \approx 9.95\).
Question 2: Using the binomial theorem, find the remainder when \(8^{100}\) is divided by 7.
\(8 = 7+1\) likhenge aur \((7+1)^{100}\) expand karenge. Aakhri term ke ilawa har term mein 7 ka factor hoga, to remainder sirf aakhri term se aayega.
\[
8^{100} = (7+1)^{100} = 7^{100} + \binom{100}{1}7^{99} + \dots + \binom{100}{99}7 + 1 = 7\cdot m + 1
\]
for some integer \(m\). Every term except the last is a multiple of 7.
Answer: The remainder is \(1\).
Question 3: Using the binomial theorem, find the remainder when \(2^{100}\) is divided by 3.
\(2 = 3-1\) likhenge. \((3-1)^{100}\) mein power 100 even hai, to aakhri term \((-1)^{100} = +1\) hogi; baqi sab terms 3 ke multiples.
\[
2^{100} = (3-1)^{100} = 3^{100} - \binom{100}{1}3^{99} + \dots - \binom{100}{99}3 + 1 = 3\cdot m + 1
\]
for some integer \(m\).
Answer: The remainder is \(1\).
Question 4: Using the binomial theorem, find which number is larger: (i) \(19^{10} + 20^{10}\) or \(21^{10}\)
\(21^{10} = (20+1)^{10}\) expand karenge. Pehle do terms ke baad wale sab terms positive hain. Dikhayenge ke extra terms ka sum \(19^{10}\) se bara hai.
\[
21^{10} = (20+1)^{10} = 20^{10} + 10\cdot 20^9 + \binom{10}{2}20^8 + \dots + 1 \gt 20^{10} + 10\cdot 20^9 + 45\cdot 20^8
\]
\[
10\cdot 20^9 + 45\cdot 20^8 = 245\cdot 20^8 \gt 361\cdot 19^8 = 19^{10}, \quad \text{since } \frac{245}{361} \approx 0.6787 \gt \left(\frac{19}{20}\right)^8 \approx 0.6634
\]
Therefore \(21^{10} \gt 20^{10} + 19^{10}\).
Answer: \(21^{10}\) is larger.
Question 4(ii): Using the binomial theorem, find which number is larger: \(29^{15} + 30^{15}\) or \(31^{15}\).
Wohi tareeqa: \(31^{15} = (30+1)^{15}\) expand karke extra terms ka sum \(29^{15}\) se compare karenge.
\[
31^{15} = (30+1)^{15} = 30^{15} + 15\cdot 30^{14} + \binom{15}{2}30^{13} + \dots + 1 \gt 30^{15} + 15\cdot 30^{14} + 105\cdot 30^{13}
\]
\[
15\cdot 30^{14} + 105\cdot 30^{13} = 555\cdot 30^{13} \gt 841\cdot 29^{13} = 29^{15}, \quad \text{since } \frac{555}{841} \approx 0.6599 \gt \left(\frac{29}{30}\right)^{13} \approx 0.6436
\]
Therefore \(31^{15} \gt 30^{15} + 29^{15}\).
Answer: \(31^{15}\) is larger.
Question 5: Using the binomial theorem, show that: (i) \(5^5 + 7^5\) is divisible by 12.
Note: the printed claim "divisible by 36" is incorrect — \(5^5+7^5 = 19932 = 12\times 1661\), and \(19932/36\) is not an integer. What the binomial expansion actually proves is divisibility by 12, shown below.
\[
5^5 + 7^5 = (6-1)^5 + (6+1)^5
\]
\[
= \left[6^5 - 5\cdot 6^4 + 10\cdot 6^3 - 10\cdot 6^2 + 5\cdot 6 - 1\right] + \left[6^5 + 5\cdot 6^4 + 10\cdot 6^3 + 10\cdot 6^2 + 5\cdot 6 + 1\right]
\]
\[
= 2\left[6^5 + 10\cdot 6^3 + 5\cdot 6\right] = 12\left[6^4 + 10\cdot 6^2 + 5\right] = 12\times 1661 = 19932
\]
Answer: \(5^5 + 7^5 = 12\times 1661\), so it is divisible by 12 (but not by 36).
Question 5(ii): Show that \(17^{15} + 13^7\) is divisible by 6.
\(17 = 18-1\) (power 15 odd, to aakhri term \(-1\)) aur \(13 = 12+1\) (aakhri term \(+1\)). Dono expansions mein baqi sab terms 6 ke multiples honge, aur \(-1+1\) cancel ho jayega.
\[
17^{15} = (18-1)^{15} = -1 + 18a, \qquad 13^7 = (12+1)^7 = 1 + 12b
\]
for some integers \(a, b\). Adding: \(17^{15} + 13^7 = 18a + 12b = 6(3a+2b)\).
Answer: \(17^{15} + 13^7 = 6(3a+2b)\), so it is divisible by 6.
Question 5(iii): Show that \(21^9 + 19^{11}\) is divisible by 20.
\(21 = 20+1\) se aakhri term \(+1\); \(19 = 20-1\) aur power 11 odd hone se aakhri term \(-1\). Baqi sab terms 20 ke multiples.
\[
21^9 = (20+1)^9 = 1 + 20a, \qquad 19^{11} = (20-1)^{11} = -1 + 20b
\]
for some integers \(a, b\). Adding: \(21^9 + 19^{11} = 20(a+b)\).
Answer: \(21^9 + 19^{11} = 20(a+b)\), so it is divisible by 20.
Question 5(iv): Examine the claim that \(31^4 + 29^6\) is divisible by 30.
Note: the printed claim is incorrect. Expanding \((30+1)^4\) and \((30-1)^6\): the first ends in \(+1\), and the second also ends in \(+1\) (power 6 is even, so \((-1)^6 = +1\)). Their sum is \(2\) more than a multiple of 30, as shown below.
\[
31^4 = (30+1)^4 = 1 + 30a, \qquad 29^6 = (30-1)^6 = 1 + 30b
\]
for some integers \(a, b\). Adding: \(31^4 + 29^6 = 2 + 30(a+b)\). In fact \(31^4 + 29^6 = 923521 + 594823321 = 595746842 = 30\times 19858228 + 2\).
Answer: \(31^4 + 29^6\) leaves remainder \(2\) on division by 30, so it is not divisible by 30 (it is even, hence divisible by 2).
Question 5(v): Show that \(101^5 + 99^7\) is divisible by 100.
\(101 = 100+1\) se aakhri term \(+1\); \(99 = 100-1\) aur power 7 odd hone se aakhri term \(-1\). Baqi sab terms 100 ke multiples.
\[
101^5 = (100+1)^5 = 1 + 100a, \qquad 99^7 = (100-1)^7 = -1 + 100b
\]
for some integers \(a, b\). Adding: \(101^5 + 99^7 = 100(a+b)\).
Answer: \(101^5 + 99^7 = 100(a+b)\), so it is divisible by 100.
Question 6: Using the binomial theorem, find the remainder when \(3^{101}\) is divided by 8.
\(3^{101} = 3\cdot 3^{100} = 3\cdot 9^{50}\) likhenge. Phir \(9 = 8+1\) rakh kar \((8+1)^{50}\) expand karenge; aakhri term \(1\) hogi aur baqi sab 8 ke multiples. Aakhir mein 3 se multiply karenge.
\[
3^{101} = 3\cdot 3^{100} = 3\cdot 9^{50} = 3(8+1)^{50} = 3\left[1 + 50\cdot 8 + \binom{50}{2}8^2 + \dots\right] = 3 + 8\cdot m
\]
for some integer \(m\).
Answer: The remainder is \(3\).
Question 7: Using the binomial theorem, find the last digit of \(32^{32}\).
Last digit ka matlab 10 se divide karne par remainder. \((30+2)^{32}\) expand karenge: aakhri term \(2^{32}\) ke ilawa har term mein 30 (is liye 10) ka factor hoga. To last digit \(2^{32}\) ki last digit hogi.
\[
32^{32} = (30+2)^{32} = 2^{32} + 32\cdot 30\cdot 2^{31} + \dots = 2^{32} + 10\cdot m
\]
for some integer \(m\). Now \(2^{32} = 4294967296\), whose last digit is 6 (powers of 2 end in 2, 4, 8, 6 cyclically, and \(32 \div 4\) leaves remainder 0).
Answer: The last digit is \(6\).
Question 8: Using the binomial theorem, show that \(7^n - 6n\) leaves remainder 1 when divided by 6, for every positive integer \(n\).
\(7 = 6+1\) likhenge. \((6+1)^n\) mein pehle do terms \(1 + 6n\) honge; baqi sab terms mein \(6^2\) ya is se bari power hogi, yani sab 6 ke multiples. \(6n\) subtract karne par sirf 1 bachega.
\[
7^n = (6+1)^n = 1 + n\cdot 6 + \binom{n}{2}6^2 + \dots + 6^n = 1 + 6n + 6\cdot m
\]
for some integer \(m\). Therefore \(7^n - 6n = 1 + 6m\).
Answer: \(7^n - 6n = 1 + 6m\), so the remainder on division by 6 is \(1\).
Question 9: Using the binomial theorem, show that \(5^n - 1\) is divisible by 4 for every positive integer \(n\).
\(5 = 4+1\) likhenge. \((4+1)^n\) mein pehla term 1 hoga aur baqi sab terms 4 ke multiples. 1 subtract karne par poora expression 4 ka multiple ban jayega.
\[
5^n = (4+1)^n = 1 + n\cdot 4 + \binom{n}{2}4^2 + \dots + 4^n = 1 + 4m
\]
for some integer \(m\). Therefore \(5^n - 1 = 4m\).
Answer: \(5^n - 1 = 4m\), so it is divisible by 4.
Question 10: Using the binomial theorem, show that \(5^n - 2^n\) is divisible by 3 for every positive integer \(n\).
\(5 = 3+2\) likhenge. \((3+2)^n\) mein aakhri term \(2^n\) hogi aur baqi sab terms 3 ke multiples. \(2^n\) subtract karne par 3 ka multiple bachega.
\[
5^n = (3+2)^n = 3^n + \binom{n}{1}3^{n-1}\cdot 2 + \dots + 2^n = 3m + 2^n
\]
for some integer \(m\). Therefore \(5^n - 2^n = 3m\).
Answer: \(5^n - 2^n = 3m\), so it is divisible by 3.
Question 11: Prove that the sum of the squares of three consecutive odd positive integers, plus 1, is divisible by 12.
Consecutive odd integers ko \(2k+1\), \(2k+3\), \(2k+5\) likhenge. Squares ka sum plus 1 expand karke dekhenge ke 12 common factor nikalta hai.
Let the three consecutive odd integers be \(2k+1\), \(2k+3\), \(2k+5\). Then:
\[
(2k+1)^2 + (2k+3)^2 + (2k+5)^2 + 1 = (4k^2+4k+1) + (4k^2+12k+9) + (4k^2+20k+25) + 1
\]
\[
= 12k^2 + 36k + 36 = 12(k^2+3k+3)
\]
Answer: The expression equals \(12(k^2+3k+3)\), so it is divisible by 12.
Question 12: A business has revenue of Rs. 10,000,000 growing at 6% per year. Use the binomial theorem to find the revenue after 4 years.
Revenue \(= P(1+r)^n = 10{,}000{,}000(1.06)^4\). \((1+0.06)^4\) ko binomial theorem se expand karke multiply karenge.
\[
(1.06)^4 = (1+0.06)^4 = 1 + 4(0.06) + 6(0.06)^2 + 4(0.06)^3 + (0.06)^4
\]
\[
= 1 + 0.24 + 0.0216 + 0.000864 + 0.00001296 = 1.26247696
\]
\[
\text{Revenue} = 10{,}000{,}000 \times 1.26247696 = 12{,}624{,}769.60
\]
Answer: Rs. 12,624,769.60.
Question 13: A person invests Rs. 2,000,000 at 10% compound interest per year. Use the binomial theorem to find the amount after 4 years.
\(A = P(1+r)^n = 2{,}000{,}000(1.1)^4\). \((1+0.1)^4\) expand karenge.
\[
(1.1)^4 = (1+0.1)^4 = 1 + 4(0.1) + 6(0.1)^2 + 4(0.1)^3 + (0.1)^4 = 1 + 0.4 + 0.06 + 0.004 + 0.0001 = 1.4641
\]
\[
A = 2{,}000{,}000 \times 1.4641 = 2{,}928{,}200
\]
Answer: Rs. 2,928,200.
Question 14: In a tournament with 8 teams, every team plays every other team once. How many matches are played in total?
Har match do teams ka aik combination hai. 8 teams mein se 2 ka selection \(\binom{8}{2}\) tareeqon se ho sakta hai.
\[
\text{Number of matches} = \binom{8}{2} = \frac{8!}{2!\,6!} = \frac{8\times 7}{2\times 1} = 28
\]
Answer: \(28\) matches.
