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Class 11 RD Sharma Solutions- Chapter 18 Binomial Theorem – Exercise 18.1

Last Updated : 19 Jul, 2021
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Question 1. Using binomial theorem, write down the expressions of the following:

(i) (2x + 3y)5

Solution:

Using binomial theorem, we have,

(2x + 3y)5 = 5C0 (2x)5 (3y)0 + 5C1 (2x)4 (3y)1 + 5C2 (2x)3 (3y)2 + 5C3 (2x)2 (3y)3 + 5C4 (2x)1 (3y)4 + 5C5 (2x)0 (3y)5

= 32x5 + 5 (16x4) (3y) + 10 (8x3) (9y)2 + 10 (4x)2 (27y)3 + 5 (2x) (81y4) + 243 y5

= 32x5 + 240x4 y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5

(ii) (2x – 3y)4

Solution:

Using binomial theorem, we have,

(2x – 3y)4 = 4C0 (2x)4 (3y)0 – 4C1 (2x)3 (3y)1 + 4C2 (2x)2 (3y)2 – 4C3 (2x)1 (3y)3 + 4C4 (2x)0 (3y)4

= 16x4 – 4 (8x3) (3y) + 6 (4x2) (9y2) – 4 (2x) (27y3) + 81y4

= 16x4 – 96x3y + 216x2y2 – 216xy3 + 81y4

(iii) (x – 1/x)6

Solution:

Using binomial theorem, we have,

(x – 1/x)6 = 6C0 x6 (1/x)0 – 6C1 x5 (1/x)1 + 6C2 x4 (1/x)2 – 6C3 x3 (1/x)3 + 6C4 x2 (1/x)4 – 6C5 x1 (1/x)5 + 6C6 (1/x)6

= x6 – 6x5 (1/x) + 15x4 (1/x2) – 20 x3 (1/x3) + 15x2 (1/x4) – 6x (1/x5) + 1/x6

= x6 – 6x4 + 15x2 – 20 + 15/x2 – 6/x4 + 1/x6

(iv) (1 – 3x)7

Solution:

Using binomial theorem, we have,

(1 – 3x)7 = 7C0 (3x)0 – 7C1 (3x)1 + 7C2 (3x)2 – 7C3 (3x)3 + 7C4 (3x)4 – 7C5 (3x)5 + 7C6 (3x)6 – 7C7 (3x)7

= 1 – 7 (3x) + 21 (9x)2 – 35 (27x3) + 35 (81x4) – 21 (243x5) + 7 (729x6) – 2187(x7)

= 1 – 21x + 189x2 – 945x3 + 2835x4 – 5103x5 + 5103x6 – 2187x7

(v) (ax – b/x)6

Solution:

Using binomial theorem, we have,

(ax – b/x)^6 = ^{6}{}{C}_0 (ax )^6 (\frac{b}{x} )^0 - ^{6}{}{C}_1 (ax )^5 (\frac{b}{x} )^1 + ^{6}{}{C}_2 (ax )^4 (\frac{b}{x} )^2 - ^{6}{}{C}_3 (ax )^3 (\frac{b}{x} )^3 +^{6}{}{C}_4 (ax )^2 (\frac{b}{x} )^4 - ^{6}{}{C}_5 (ax )^1 (\frac{b}{x} )^5 + ^{6}{}{C}_6 (ax )^0 (\frac{b}{x} )^6

= a^6 x^6 - 6 a^5 x^5 \times \frac{b}{x} + 15 a^4 x^4 \times \frac{b^2}{x^2} - 20 a^3 b^3 \times \frac{b^3}{x^3} + 15 a^2 x^2 \times \frac{b^4}{x^4} - 6ax \times \frac{b^5}{x^5} + \frac{b^6}{x^6}

= a^6 x^6 - 6 a^5 x^4 b + 15 a^4 x^2 b^2 - 20 a^3 b^3 + 15\frac{a^2 b^4}{x^2} - 6\frac{a b^5}{x^4} + \frac{b^6}{x^6}

(vi) \left(\frac{\sqrt{x}}{a} - \sqrt{\frac{a}{x}} \right)^6

Solution:

Using binomial theorem, we have,

\left( \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}} \right)^6 = ^{6}{}{C}_0 \left( \sqrt{\frac{x}{a}} \right)^6 \left( \sqrt{\frac{a}{x}} \right)^0 - ^{6}{}{C}_1 \left( \sqrt{\frac{x}{a}} \right)^5 \left( \sqrt{\frac{a}{x}} \right)^1 + ^{6}{}{C}_2 \left( \sqrt{\frac{x}{a}} \right)^4 \left( \sqrt{\frac{a}{x}} \right)^2 - ^{6}{}{C}_3 \left( \sqrt{\frac{x}{a}} \right)^3 \left( \sqrt{\frac{a}{x}} \right)^3 +^{6}{}{C}_4 \left( \sqrt{\frac{x}{a}} \right)^2 \left( \sqrt{\frac{a}{x}} \right)^4 -^{6}{}{C}_5 \left( \sqrt{\frac{x}{a}} \right)^1 \left( \sqrt{\frac{a}{x}} \right)^5 + ^{6}{}{C}_6 \left( \sqrt{\frac{x}{a}} \right)^0 \left( \sqrt{\frac{a}{x}} \right)^6

= \frac{x^3}{a^3} - 6\frac{x^2}{a^2} + 15\frac{x}{a} - 20 + 15\frac{a}{x} - 6\frac{a^2}{x^2} + \frac{a^3}{x^3}

(vii) \left( \sqrt[3]{x} - \sqrt[3]{a} \right)^6

Solution:

Using binomial theorem, we have,

\left( \sqrt[3]{x} - \sqrt[3]{a} \right)^6 = ^{6}{}{C}_0 (\sqrt[3]{x} )^6 (\sqrt[3]{a} )^0 -^{6}{}{C}_1 (\sqrt[3]{x} )^5 (\sqrt[3]{a} )^1 +^{6}{}{C}_2 (\sqrt[3]{x} )^4 (\sqrt[3]{a} )^2 -^{6}{}{C}_3 (\sqrt[3]{x} )^3 (\sqrt[3]{a} )^3 +^{6}{}{C}_4 (\sqrt[3]{x} )^2 (\sqrt[3]{a} )^4 -^{6}{}{C}_5 (\sqrt[3]{x} )^1 (\sqrt[3]{a} )^5 + ^{6}{}{C}_6 (\sqrt[3]{x} )^0 (\sqrt[3]{a} )^6

= x^2 - 6 x^{5/3} a^{1/3} + 15 x^{4/3} a^{2/3} - 20xa + 15 x^{2/3} a^{4/3} - 6 x^{1/3} a^{5/3} + a^2

(viii) (1 + 2x – 3x2)5

Solution:

Using binomial theorem, we have,

(1 + 2x – 3x2)5 = 5C0 (1 + 2x)5 (3x2)0 – 5C1 (1 + 2x)4 (3x2)1 + 5C2 (1 + 2x)3 (3x2)2 – 5C3 (1 + 2x)2 (3x2)3 + 5C4 (1 + 2x)1 (3x2)4 – 5C5 (1 + 2x)0 (3x2)5

= (1 + 2x)5 – 5(1 + 2x)4 (3x2) + 10 (1 + 2x)3 (9x4) – 10 (1 + 2x)2 (27x6) + 5 (1 + 2x) (81x8) – 243x10

= 5C0 (2x)0 + 5C1 (2x)1 + 5C2 (2x)2 + 5C3 (2x)3 + 5C4 (2x)4 + 5C5 (2x)5 – 15x2 [4C0 (2x)0 + 4C1 (2x)1 + 4C2 (2x)2 + 4C3 (2x)3 + 4C4 (2x)4] + 90x4 [1 + 8x3 + 6x + 12x2] – 270x6(1 + 4x2 + 4x) + 405x8 + 810x9 – 243x10

= 1 + 10x + 40x2 + 80x3 + 80x4 + 32x5 – 15x2 – 120x3 – 3604 – 480x5 – 240x6 + 90x4 + 720x7 + 540x5 + 1080x6 – 270x6 – 1080x8 – 1080x7 + 405x8 + 810x9 – 243x10

= 1 + 10x + 25x2 – 40x3 – 190x4 + 92x5 + 570x6 – 360x7 – 675x8 + 810x9 – 243x10

(ix) \left( x + 1 - \frac{1}{x} \right)^3

Solution:

Using binomial theorem, we have,

(x + 1 – 1/x)3 = 3C0 (x + 1)3 (1/x)0 – 3C1(x + 1)2(1/x)1 + 3C2(x + 1)1(1/x)2 – 3C3 (x + 1)0 (1/x)3

= (x + 1 )^3 - 3(x + 1 )^2 \times \frac{1}{x} + 3\frac{x + 1}{x^2} - \frac{1}{x^3}

= x^3 + 1 + 3x + 3 x^2 - \frac{3 x^2 + 3 + 6x}{x} + 3\frac{x + 1}{x^2} - \frac{1}{x^3}

= x^3 + 1 + 3x + 3 x^2 - 3x - \frac{3}{x} - 6 + \frac{3}{x} + \frac{3}{x^2} - \frac{1}{x^3}

= x^3 + 3 x^2 - 5 + \frac{3}{x^2} - \frac{1}{x^3}

(x) (1 – 2x + 3x2)3

Solution:

Using binomial theorem, we have,

(1 – 2x + 3 x2)3 = 3C0 (1 – 2x)3 + 3C1 (1 – 2x)2 (3x2) + 3C2 (1 – 2x)(3x2)2 + 3C3 (3x2)3

= (1 – 2x)3 + 9x2 (1 – 2x)2 + 27x4 (1 – 2x) + 27x6

= 1 – 8x3 + 12x2 – 6x + 9x2 (1 + 4x2 – 4x) + 27x4 – 54x5 + 27x6 

= 1 – 8x3 + 12x2 – 6x + 9x2 + 36x4 – 36x3 + 27x4 – 54x5 + 27x6

= 1 – 6x + 21x2 – 44x3 + 63x4 – 54x5 + 27x6 

Question 2. Evaluate the following:

(i) \left( \sqrt{x + 1} + \sqrt{x - 1} \right)^6 + \left( \sqrt{x + 1} - \sqrt{x - 1} \right)^6

Solution:

Using binomial theorem, we have,

\left( \sqrt{x + 1} + \sqrt{x - 1} \right)^6 + \left( \sqrt{x + 1} - \sqrt{x - 1} \right)^6  = 2[ ^{6}{}{C}_0 (\sqrt{x + 1} )^6 (\sqrt{x - 1} )^0 + ^{6}{}{C}_2 (\sqrt{x + 1} )^4 (\sqrt{x - 1} )^2 +^{6}{}{C}_4 (\sqrt{x + 1} )^2 (\sqrt{x - 1} )^4 + ^{6}{}{C}_6 (\sqrt{x + 1} )^0 (\sqrt{x - 1} )^6 ]

= 2 [(x + 1)3 + 15(x + 1)2 (x – 1) + 15(x + 1)(x – 1 )2 + (x – 1 )3]

= 2 [x3 + 1 + 3x + 3 x2 + 15( x2 + 2x + 1)(x – 1) + 15(x + 1)( x2 + 1 – 2x) + x3 – 1 + 3x – 3 x2]

= 2 [2 x3 + 6x + 15 x3 – 15 x2 + 30 x2 – 30x + 15x – 15 + 15 x3 + 15 x2 – 30 x2 – 30x + 15x + 15]

= 2 [32 x3 – 24x]

= 16x [4x2 – 3]

(ii) \left( x + \sqrt{x^2 - 1} \right)^6 + \left( x - \sqrt{x^2 - 1} \right)^6

Solution:

Using binomial theorem, we have,

(x + \sqrt{x^2 - 1} )^6 + (x - \sqrt{x^2 - 1} )^6 = 2[ ^ {6}{}{C}_0 x^6 (\sqrt{x^2 - 1} )^0 +^{6}{}{C}_2 x^4 (\sqrt{x^2 - 1} )^2 +^{6}{}{C}_4 x^2 (\sqrt{x^2 - 1} )^4 + ^{6}{}{C}_6 x^0 (\sqrt{x^2 - 1} )^6 ]

= 2 [x6 + 15 x4 ( x2 – 1) + 15 x2 ( x2 – 1 )2 + ( x2 – 1 )3]

= 2 [x6 + 15 x6 – 15 x4 + 15 x2 ( x4 – 2 x2 + 1) + ( x6 – 1 + 3 x2 – 3 x4)]

= 2 [x6 + 15 x6 – 15 x4 + 15 x6 – 30 x4 + 15 x2 + x6 – 1 + 3 x2 – 3 x4]

= 64 x6 – 96 x4 + 36 x2 – 2

(iii) (1+2\sqrt{x})^5+(1-2\sqrt{x})^5

Solution:

Using binomial theorem, we have,

(1+2\sqrt{x})^5+(1-2\sqrt{x})^5 = 2 [5C0 (2√x)0 + 5C2 (2√x)2 + 5C4 (2√x)4]

= 2 [1 + 10 (4x) + 5 (16x2)]

= 2 [1 + 40x + 80x2]

(iv) (\sqrt{2}+1)^6+(\sqrt{2}-1)^6

Solution:

Using binomial theorem, we have,

(\sqrt{2}+1)^6+(\sqrt{2}-1)^6 = 2 [6C0 (√2)6 + 6C2 (√2)4 + 6C4 (√2)2 + 6C6 (√2)0]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [99]

= 198

(v) (3+\sqrt{2})^5+(3-\sqrt{2})^5

Solution:

Using binomial theorem, we have,

(3+\sqrt{2})^5+(3-\sqrt{2})^5  = 2 [5C1 (34) (√2)1 + 5C3 (32) (√2)3 + 5C5 (30) (√2)5]

= 2 [5 (81) (√2) + 10 (9) (2√2) + 4√2]

= 2√2 (405 + 180 + 4)

= 1178√2

(vi) (2+\sqrt{3})^7+(2-\sqrt{3})^7

Solution:

Using binomial theorem, we have,

(2+\sqrt{3})^7+(2-\sqrt{3})^7  = 2 [7C0 (27) (√3)0 + 7C2 (25) (√3)2 + 7C4 (23) (√3)4 + 7C6 (21) (√3)6]

= 2 [128 + 21 (32)(3) + 35(8)(9) + 7(2)(27)]

= 2 [128 + 2016 + 2520 + 378]

= 2 [5042]

= 10084

(vii) (\sqrt{3}+1)^5+(\sqrt{3}-1)^5

Solution:

Using binomial theorem, we have,

(\sqrt{3}+1)^5+(\sqrt{3}-1)^5  = 2 [5C1 (√3)4 + 5C3 (√3)2 + 5C5 (√3)0]

= 2 [5 (9) + 10 (3) + 1]

= 2 [76]

= 152

(viii) (0.99)5 + (1.01)5

Solution:

Using binomial theorem, we have,

(0.99)5 + (1.01)5 = (1 – 0.01)5 + (1 + 0.01)5

= 2 [5C0 (0.01)0 + 5C2 (0.01)2 + 5C4 (0.01)4]

= 2 [1 + 10 (0.0001) + 5 (0.00000001)]

= 2 [1.00100005]

= 2.0020001

(ix) (\sqrt{3}+\sqrt{2})^6+(\sqrt{3}-\sqrt{2})^6

Solution:

Using binomial theorem, we have,

(\sqrt{3}+\sqrt{2})^6+(\sqrt{3}-\sqrt{2})^6  = 2 [6C1 (√3)5 (√2)1 + 6C3 (√3)3 (√2)3 + 6C5 (√3)1 (√2)5]

= 2 [6 (9√3) (√2) + 20 (3√3) (2√2) + 6 (√3) (4√2)]

= 2 [√6 (54 + 120 + 24)]

= 396 √6

(x) \left\{ a^2 + \sqrt{a^2 - 1} \right\}^4 + \left\{ a^2 - \sqrt{a^2 - 1} \right\}^4

Solution:

Using binomial theorem, we have,

\left\{ a^2 + \sqrt{a^2 - 1} \right\}^4 + \left\{ a^2 - \sqrt{a^2 - 1} \right\}^4 = 2[ ^{4}{}{C}_0 ( a^2 )^4 (\sqrt{a^2 - 1} )^0 +^{4}{}{C}_2 ( a^2 )^2 (\sqrt{a^2 - 1} )^2 + ^{4}{}{C}_4 ( a^2 )^0 (\sqrt{a^2 - 1} )^4 ]

= 2[ a8 + 6 a4 (a2 – 1) + ( a2 – 1 )2]

= 2[a8 + 6 a6 – 6 a4 + a4 + 1 – 2 a2]

= 2 a8 + 12 a6 – 10 a4 – 4 a2 + 2

Question 3. Find (a + b)4 – (a – b)4. Hence, evaluate (√3 + √2)4 – (√3 – √2)4.

Solution:

We are given,

(a + b)4 – (a – b)4 = 2 [4C1 a3b1 + 4C3 a1b3]

= 2 [4a3b + 4ab3]

= 8 (a3b + ab3)

Now,

(√3 + √2)4 – (√3 – √2)4 = 8 (a3b + ab3)

= 8 [(√3)3 (√2) + (√3) (√2)3]

= 8 [(3√6) + (2√6)]

= 8 (5√6)

= 40√6

Question 4. Find (x + 1)6 + (x – 1)6. Hence, or otherwise evaluate (√2 + 1)6 + (√2 – 1)6.

Solution:

We are given,

(x + 1)6 + (x – 1)6 = 2 [6C0 x6 + 6C2 x4 + 6C4 x2 + 6C6 x0]

= 2 [x6 + 15x4 + 15x2 + 1]

Now,

(√2 + 1)6 + (√2 – 1)6

So consider, x = √2 then we get,

(√2 + 1)6 + (√2 – 1)6 = 2 [x6 + 15x4 + 15x2 + 1]

= 2 [(√2)6 + 15 (√2)4 + 15 (√2)2 + 1]

= 2 [8 + 15 (4) + 15 (2) + 1]

= 2 [8 + 60 + 30 + 1]

= 198

Question 5. Using binomial theorem evaluate each of the following:

(i) (96)3

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(96)3 = (100 – 4)3

= 3C0 (100)3 (4)0 – 3C1 (100)2 (4)1 + 3C2 (100)1 (4)2 – 3C3 (100)0 (4)3

= 1000000 – 120000 + 4800 – 64

= 884736

(ii) (102)5

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(102)5 = (100 + 2)5

= 5C0 (100)5 (2)0 + 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 + 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 + 5C5 (100)0 (2)5

= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

(iii) (101)4

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(101)4 = (100 + 1)4

= 4C0 (100)4 + 4C1 (100)3 + 4C2 (100)2 + 4C3 (100)1 + 4C4 (100)0

= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401

(iv) (98)5

Solution:

On expressing the given expression as two different entities and applying the binomial theorem, we get,

(98)5 = (100 – 2)5

= 5C0 (100)5 (2)0 – 5C1 (100)4 (2)1 + 5C2 (100)3 (2)2 – 5C3 (100)2 (2)3 + 5C4 (100)1 (2)4 – 5C5 (100)0 (2)5

= 10000000000 – 1000000000 + 40000000 – 800000 + 8000 – 32

= 9039207968

Question 6. Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.

Solution:

We are given,

23n – 7n – 1 = 8n – 7n – 1

= (1 + 7)n – 7n – 1

= nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + .… + nCn (7)n  â€“ 7n – 1

= 1 + 7n + 72 [ nC2 + nC3 (71) + nC4 (72) + … + nCn (7)n-2] – 7n – 1

= 49 [ nC2 + nC3 (71) + nC4 (72) + … + nCn (7)n-2] , which is divisible by 49.

Therefore, 23n – 1 – 7n is divisible by 49.

Hence proved.

Question 7. Using the binomial theorem, prove that 32n+2 – 8n – 9 is divisible by 64, where n ∈ N.

Solution:

We are given,

32n+2 – 8n – 9 = 32(n+1) – 8n – 9

= 9n+1 – 8n – 9

= (1 + 8)n+1 – 8n – 9

= n+1C0 + n+1C1 (8)1 + n+1C2 (8)2 + n+1C3 (8)3 + n+1C4 (8)2 + n+1C5 (8)1 + .… + n+1Cn+1 (8)n+1 – 8n – 9

= 1 + 8(n+1) + 82 [ n+1C2 + n+1C3 (81) + n+1C4 (82) + … + n+1Cn+1 (8)n-1] – 8n – 9

= 8n + 9 + 64 [ n+1C2 + n+1C3 (81) + n+1C4 (82) + … + n+1Cn+1 (8)n-1] – 8n – 9

= 64 [ n+1C2 + n+1C3 (81) + n+1C4 (82) + … + n+1Cn+1 (8)n-1], which is divisible by 64.

Therefore, 32n+2 – 8n – 9 is divisible by 64.

Hence proved.

Question 8. If n is a positive integer, prove that 33n – 26n – 1 is divisible by 676.

Solution:

We are given,

33n – 26n – 1 = (33)n – 26n – 1

= 27n – 26n – 1

= (1 + 26)n – 26n – 1

= nC0 + nC1 (26)1 + nC2 (26)2 + nC3 (26)3 + nC4 (26)2 + nC5 (26)1 + .… + nCn (26)n  â€“ 26n – 1

= 1 + 26n + 262 [ nC2 + nC3 (261) + nC4 (262) + … + nCn (26)n-2] – 26n – 1

= 676 [ nC2 + nC3 (261) + nC4 (262) + … + nCn (26)n-2] , which is divisible by 676.

Therefore, 33n – 26n – 1 is divisible by 676.

Hence proved.

Question 9. Using binomial theorem, indicate which is larger (1.1)10000 or 1000.

Solution:

We have,

(1.1)10000 = (1 + 0.1)10000

= 10000C0 + 10000C1 (0.1)1 + 10000C2 (0.1)2 + .… + 10000C10000 (0.1)10000

= 1 + (10000) (0.1) + other positive terms

= 1 + 1000 + other positive terms

= 1001 + other positive terms

Therefore, (1.1)10000 is larger than 1000.

Question 10. Using binomial theorem, determine which number is larger, (1.2)4000 or 800?

Solution:

We have,

(1.2)4000 = (1 + 0.2)4000

= 4000C0 + 4000C1 (0.2)1 + 4000C2 (0.2)2 + .… + 4000C4000 (0.2)4000

= 1 + (4000) (0.2) + other positive terms

= 1 + 800 + other positive terms

= 801 + other positive terms

Therefore, (1.2)4000 is larger than 800.

Question 11. Find the value of (1.01)10 + (1−0.01)10 correct to 7 places of decimal.

Solution:

We have, 

(1.01)10 + (1−0.01)10 = (1+0.01)10 + (1−0.01)10

= \left(^{10}C_1+^{10}C_2(\frac{1}{10})^2+\hspace{0.1cm}^{10}C_3(\frac{1}{10})^3+...+^{10}C_{10}(\frac{1}{10})^{10}\right)+\left(^{10}C_1-^{10}C_2(\frac{1}{10})^2+\hspace{0.1cm}^{10}C_3(\frac{1}{10})^3-^{10}C_4(\frac{1}{10})^4+...-^{10}C_{10}(\frac{1}{10})^{10}\right)

= 2\left(^{10}C_1+^{10}C_3(\frac{1}{10})^3+^{10}C_5(\frac{1}{10})^5+^{10}C_7(\frac{1}{10})^7+^{10}C_9(\frac{1}{10})^9\right)

= 2\left(10+\frac{10!}{3!7!}(\frac{1}{1000})+\frac{10!}{5!5!}(\frac{1}{10^5})+\frac{10!}{7!3!}(\frac{1}{10^7})+\frac{10!}{1!9!}(\frac{1}{10^9})\right)

= 2\left(10+\frac{10×9×8}{3×2×1×1000}+\frac{10×9×8×7×6}{5×4×3×2×1×10^5}+\frac{10×9×8}{3×2×1×10^7}+\frac{1}{10^8}\right)

= 2.0090042

Question 12. Show that 24n+4 − 15n − 16, where n ∈ N is divisible by 225.

Solution:

We have,

24n+4 − 15n − 16 = 24(n+1) − 15n − 16

= 16n+1 − 15n − 16

= (1 + 15)n+1 − 15n − 16

= n+1C0 + n+1C1 (15)1 + n+1C2 (15)2 + n+1C3 (15)3 + n+1C4 (15)2 + n+1C5 (15)1 + .… + n+1Cn+1 (15)n+1 − 15n − 16

= 1 + 15(n+1) + 152 [ n+1C2 + n+1C3 (151) + n+1C4 (152) + … + n+1Cn+1 (15)n-1] – 15n – 16

= 15n + 16 + 225 [ n+1C2 + n+1C3 (151) + n+1C4 (152) + … + n+1Cn+1 (15)n-1] – 15n – 16

= 225 [ n+1C2 + n+1C3 (151) + n+1C4 (152) + … + n+1Cn+1 (15)n-1] , which is divisible by 225.

Therefore, 24n+4 − 15n − 16 is divisible by 225.

Hence proved.



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Class 11 RD Sharma Solution - Chapter 18 Binomial Theorem- Exercise 18.2 | Set 1
Class 11 NCERT Solutions- Chapter 8 Binomial Theorem - Exercise 8.1
Class 11 NCERT Solutions - Chapter 8 Binomial Theorem - Exercise 8.2
Class 12 RD Sharma Solutions - Chapter 33 Binomial Distribution - Exercise 33.1 | Set 1
Class 12 RD Sharma Solutions- Chapter 33 Binomial Distribution - Exercise 33.2 | Set 1
Class 12 RD Sharma Solutions - Chapter 33 Binomial Distribution - Exercise 33.2 | Set 2
Class 12 RD Sharma Solutions - Chapter 33 Binomial Distribution - Exercise 33.1 | Set 2
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