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Binomial Theorem

CBSE · Class 11 · Mathematics

NCERT Solutions for Binomial Theorem — CBSE Class 11 Mathematics.

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20 Questions Solved · 2 Sections

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EXERCISE 7.1

1(1−2x)5(1 - 2x)^5Show solution
Using the binomial theorem,
(1−2x)5=∑r=055Cr(1)5−r(−2x)r. (1-2x)^5=\sum_{r=0}^{5} {^5C_r}(1)^{5-r}(-2x)^r.
Now compute term by term:
=1−5(2x)+10(2x)2−10(2x)3+5(2x)4−(2x)5. =1-5(2x)+10(2x)^2-10(2x)^3+5(2x)^4-(2x)^5.
So,
(1−2x)5=1−10x+40x2−80x3+80x4−32x5. (1-2x)^5=1-10x+40x^2-80x^3+80x^4-32x^5.

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2(2x−x2)5\left(\frac{2}{x} - \frac{x}{2}\right)^5Show solution
Using the binomial theorem,
(2x−x2)5=∑r=055Cr(2x)5−r(−x2)r. \left(\frac{2}{x}-\frac{x}{2}\right)^5=\sum_{r=0}^{5} {^5C_r}\left(\frac{2}{x}\right)^{5-r}\left(-\frac{x}{2}\right)^r.
Compute each term:
=(2x)5−5(2x)4(x2)+10(2x)3(x2)2 =\left(\frac{2}{x}\right)^5-5\left(\frac{2}{x}\right)^4\left(\frac{x}{2}\right)+10\left(\frac{2}{x}\right)^3\left(\frac{x}{2}\right)^2
−10(2x)2(x2)3+5(2x)(x2)4−(x2)5. -10\left(\frac{2}{x}\right)^2\left(\frac{x}{2}\right)^3+5\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)^4-\left(\frac{x}{2}\right)^5.
Simplifying,
=32x5−40x3+20x−5x+5x316−x532. =\frac{32}{x^5}-\frac{40}{x^3}+\frac{20}{x}-5x+\frac{5x^3}{16}-\frac{x^5}{32}.

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3(2x−3)6(2x - 3)^6Show solution
Using the binomial theorem,
(2x−3)6=∑r=066Cr(2x)6−r(−3)r. (2x-3)^6=\sum_{r=0}^{6} {^6C_r}(2x)^{6-r}(-3)^r.
Now expand term by term:
(2x)6−6(2x)5(3)+15(2x)4(32)−20(2x)3(33)+15(2x)2(34)−6(2x)(35)+36. (2x)^6-6(2x)^5(3)+15(2x)^4(3^2)-20(2x)^3(3^3)+15(2x)^2(3^4)-6(2x)(3^5)+3^6.
So,
(2x−3)6=64x6−576x5+2160x4−4320x3+4860x2−2916x+729. (2x-3)^6=64x^6-576x^5+2160x^4-4320x^3+4860x^2-2916x+729.

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4(x3+1x)5\left(\frac{x}{3} + \frac{1}{x}\right)^5Show solution
Using the binomial theorem,
(x3+1x)5=∑r=055Cr(x3)5−r(1x)r. \left(\frac{x}{3}+\frac{1}{x}\right)^5=\sum_{r=0}^{5}{^5C_r}\left(\frac{x}{3}\right)^{5-r}\left(\frac{1}{x}\right)^r.
Compute the terms:
=(x3)5+5(x3)41x+10(x3)31x2+10(x3)21x3+5(x3)1x4+1x5. =\left(\frac{x}{3}\right)^5+5\left(\frac{x}{3}\right)^4\frac{1}{x}+10\left(\frac{x}{3}\right)^3\frac{1}{x^2}+10\left(\frac{x}{3}\right)^2\frac{1}{x^3}+5\left(\frac{x}{3}\right)\frac{1}{x^4}+\frac{1}{x^5}.
Simplifying,
(x3+1x)5=x5243+5x381+10x27+103x+5x3+1x5. \left(\frac{x}{3}+\frac{1}{x}\right)^5=\frac{x^5}{243}+\frac{5x^3}{81}+\frac{10x}{27}+\frac{10}{3x}+\frac{5}{x^3}+\frac{1}{x^5}.

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5(x+1x)6\left(x + \frac{1}{x}\right)^6Show solution
Using the binomial theorem,
(x+1x)6=∑r=066Crx6−r(1x)r. \left(x+\frac{1}{x}\right)^6=\sum_{r=0}^{6}{^6C_r}x^{6-r}\left(\frac{1}{x}\right)^r.
So,
=x6+6x4+15x2+20+15x−2+6x−4+x−6. = x^6+6x^4+15x^2+20+15x^{-2}+6x^{-4}+x^{-6}.
Hence,
(x+1x)6=x6+6x4+15x2+20+15x2+6x4+1x6. \left(x+\frac{1}{x}\right)^6=x^6+6x^4+15x^2+20+\frac{15}{x^2}+\frac{6}{x^4}+\frac{1}{x^6}.

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6(96)3(96)^3Show solution

963=(100−4)3 96^3=(100-4)^3
Using (a−b)3=a3−3a2b+3ab2−b3(a-b)^3=a^3-3a^2b+3ab^2-b^3,
(100−4)3=1003−3⋅1002⋅4+3⋅100⋅42−43. (100-4)^3=100^3-3\cdot100^2\cdot4+3\cdot100\cdot4^2-4^3.
=1000000−120000+4800−64=884736. =1000000-120000+4800-64=884736.

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7(102)5(102)^5Show solution

1025=(100+2)5 102^5=(100+2)^5
Using the binomial theorem,
(100+2)5=1005+5⋅1004⋅2+10⋅1003⋅22+10⋅1002⋅23+5⋅100⋅24+25. (100+2)^5=100^5+5\cdot100^4\cdot2+10\cdot100^3\cdot2^2+10\cdot100^2\cdot2^3+5\cdot100\cdot2^4+2^5.
Now calculate:
1005=10000000000, 100^5=10000000000,
5⋅1004⋅2=1000000000, 5\cdot100^4\cdot2=1000000000,
10⋅1003⋅4=40000000, 10\cdot100^3\cdot4=40000000,
10⋅1002⋅8=800000, 10\cdot100^2\cdot8=800000,
5⋅100⋅16=8000, 5\cdot100\cdot16=8000,
25=32. 2^5=32.
Adding,
10000000000+1000000000+40000000+800000+8000+32=11040808032. 10000000000+1000000000+40000000+800000+8000+32=11040808032.
So the value is 1104080803211040808032.

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8(101)4(101)^4Show solution

1014=(100+1)4 101^4=(100+1)^4
Using (a+b)4=a4+4a3b+6a2b2+4ab3+b4(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4,
(100+1)4=1004+4⋅1003+6⋅1002+4⋅100+1. (100+1)^4=100^4+4\cdot100^3+6\cdot100^2+4\cdot100+1.
=100000000+4000000+60000+400+1=104060401. =100000000+4000000+60000+400+1=104060401.

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9(99)5(99)^5Show solution

995=(100−1)5 99^5=(100-1)^5
Using the binomial theorem,
(100−1)5=1005−5⋅1004+10⋅1003−10⋅1002+5⋅100−1. (100-1)^5=100^5-5\cdot100^4+10\cdot100^3-10\cdot100^2+5\cdot100-1.
Now compute:
1005=10000000000, 100^5=10000000000,
5⋅1004=500000000, 5\cdot100^4=500000000,
10⋅1003=10000000, 10\cdot100^3=10000000,
10⋅1002=100000, 10\cdot100^2=100000,
5⋅100=500. 5\cdot100=500.
So,
10000000000−500000000+10000000−100000+500−1=950990049. 10000000000-500000000+10000000-100000+500-1=950990049.

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10Using Binomial Theorem, indicate which number is larger (1.1)10000(1.1)^{10000} or 1000.Show solution
Using the binomial theorem,
(1.1)10000=(1+0.1)10000. (1.1)^{10000}=(1+0.1)^{10000}.
All terms in the expansion are positive, so
(1+0.1)10000>1+10000(0.1)=1001. (1+0.1)^{10000} > 1 + 10000(0.1)=1001.
Therefore,
(1.1)10000>1000. (1.1)^{10000} > 1000.
So (1.1)^{10000} is larger.

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11Find (a+b)4−(a−b)4(a + b)^4 - (a - b)^4. Hence, evaluate (3+2)4−(3−2)4(\sqrt{3} + \sqrt{2})^4 - (\sqrt{3} - \sqrt{2})^4.
12Find (x+1)6+(x−1)6(x + 1)^6 + (x - 1)^6. Hence or otherwise evaluate (2+1)6+(2−1)6(\sqrt{2} + 1)^6 + (\sqrt{2} - 1)^6.
13Show that 9n+1−8n−99^{n+1} - 8n - 9 is divisible by 64, whenever nn is a positive integer.
14Prove that ∑r=0n3rnCr=4n\sum_{r=0}^n 3^r {}^nC_r = 4^n.

Miscellaneous Exercise on Chapter 7

1If aa and bb are distinct integers, prove that a−ba - b is a factor of an−bna^n - b^n, whenever nn is a positive integer.
2Evaluate (3+2)6−(3−2)6(\sqrt{3} + \sqrt{2})^6 - (\sqrt{3} - \sqrt{2})^6.
3Find the value of (a2+a2−1)4+(a2−a2−1)4\left(a^2 + \sqrt{a^2 - 1}\right)^4 + \left(a^2 - \sqrt{a^2 - 1}\right)^4.
4Find an approximation of (0.99)5(0.99)^5 using the first three terms of its expansion.
5Expand using Binomial Theorem (1+x2−2x)4,x≠0\left(1 + \frac{x}{2} - \frac{2}{x}\right)^4, x \neq 0.
6Find the expansion of (3x2−2ax+3a2)3(3x^2 - 2ax + 3a^2)^3 using binomial theorem.

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