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Binomial Theorem

Class 11 · Maths · Chapter 7 · Miscellaneous Exercise

NCERT Solutions for Class 11 Maths Chapter 7 Binomial Theorem with fully worked steps for expansions, powers, terms, divisibility and remainder proofs

As per current NCERT edition.

Miscellaneous Exercise

Q1. If 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. [Hint write an=(a−b+b)na^n=(a-b+b)^n and expand]

Sol. To prove that a−ba-b is a factor of an−bna^n-b^n, we have to show that

an−bn=(a−b)k a^n-b^n=(a-b)k

where kk is an integer.

We have,

an=(a−b+b)n=[(a−b)+b]n \begin{array}{l} \displaystyle a^n=(a-b+b)^n \\[0.4em] \displaystyle =[(a-b)+b]^n \end{array}

By the binomial theorem,

(a+b)n=nC0an+nC1an−1b+nC2an−2b2+⋯+nCnbn\textcolor{#1B5E20}{ (a+b)^n={}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\dots+{}^nC_nb^n }

Replacing aa by (a−b)(a-b),

[(a−b)+b]n=nC0(a−b)n+nC1(a−b)n−1b+nC2(a−b)n−2b2+⋯+nCnbnan=(a−b)n+nC1(a−b)n−1b+⋯+bnan−bn=(a−b)n+nC1(a−b)n−1b+⋯+nCn−1(a−b)bn−1 \begin{array}{l} \displaystyle [(a-b)+b]^n={}^nC_0(a-b)^n+{}^nC_1(a-b)^{n-1}b+{}^nC_2(a-b)^{n-2}b^2+\dots+{}^nC_nb^n \\[0.4em] \displaystyle a^n=(a-b)^n+{}^nC_1(a-b)^{n-1}b+\dots+b^n \\[0.4em] \displaystyle a^n-b^n=(a-b)^n+{}^nC_1(a-b)^{n-1}b+\dots+{}^nC_{n-1}(a-b)b^{n-1} \end{array}

Taking (a−b)(a-b) common,

an−bn=(a−b)[(a−b)n−1+nC1(a−b)n−2b+⋯+nCn−1bn−1]an−bn=(a−b)k \begin{array}{l} \displaystyle a^n-b^n=(a-b)\mathopen{[}(a-b)^{n-1}+{}^nC_1(a-b)^{n-2}b+\dots+{}^nC_{n-1}b^{n-1}\mathclose{]} \\[0.4em] \displaystyle a^n-b^n=(a-b)k \end{array}

where kk is an integer.

∴ a−ba-b is a factor of an−bna^n-b^n.

Q2. Evaluate (3+2)6−(3−2)6\left(\sqrt{3}+\sqrt{2}\right)^6-\left(\sqrt{3}-\sqrt{2}\right)^6.

Sol. It is in the form (a+b)6−(a−b)6(a+b)^6-(a-b)^6.

Here, a=3a=\sqrt{3} and b=2b=\sqrt{2}.

By the binomial theorem,

(a+b)n=nC0an+nC1an−1b+nC2an−2b2+⋯+nCnbn\textcolor{#1B5E20}{ (a+b)^n={}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\dots+{}^nC_nb^n }
(a+b)6=6C0a6+6C1a5b+6C2a4b2+6C3a3b3+6C4a2b4+6C5ab5+6C6b6(a−b)6=6C0a6−6C1a5b+6C2a4b2−6C3a3b3+6C4a2b4−6C5ab5+6C6b6 \begin{array}{l} \displaystyle (a+b)^6={}^6C_0a^6+{}^6C_1a^5b+{}^6C_2a^4b^2+{}^6C_3a^3b^3+{}^6C_4a^2b^4+{}^6C_5ab^5+{}^6C_6b^6 \\[0.4em] \displaystyle (a-b)^6={}^6C_0a^6-{}^6C_1a^5b+{}^6C_2a^4b^2-{}^6C_3a^3b^3+{}^6C_4a^2b^4-{}^6C_5ab^5+{}^6C_6b^6 \end{array}

Subtracting,

(a+b)6−(a−b)6=(6C0a6+6C1a5b+6C2a4b2+6C3a3b3+6C4a2b4+6C5ab5+6C6b6)−(6C0a6−6C1a5b+6C2a4b2−6C3a3b3+6C4a2b4−6C5ab5+6C6b6)=2(6C1a5b+6C3a3b3+6C5ab5)=2(6a5b+20a3b3+6ab5) \begin{array}{l} \displaystyle (a+b)^6-(a-b)^6 \\[0.4em] \displaystyle =\mathopen{(}{}^6C_0a^6+{}^6C_1a^5b+{}^6C_2a^4b^2+{}^6C_3a^3b^3+{}^6C_4a^2b^4+{}^6C_5ab^5+{}^6C_6b^6\mathclose{)} \\[0.4em] \displaystyle \mathrel{-}\mathopen{(}{}^6C_0a^6-{}^6C_1a^5b+{}^6C_2a^4b^2-{}^6C_3a^3b^3+{}^6C_4a^2b^4-{}^6C_5ab^5+{}^6C_6b^6\mathclose{)} \\[0.4em] \displaystyle =2({}^6C_1a^5b+{}^6C_3a^3b^3+{}^6C_5ab^5) \\[0.4em] \displaystyle =2(6a^5b+20a^3b^3+6ab^5) \end{array}

Putting a=3a=\sqrt{3} and b=2b=\sqrt{2},

(3+2)6−(3−2)6=2[6(3)52+20(3)3(2)3+63(2)5]=2[6×93×2+20×33×22+63×42]=2(546+1206+246)=2(1986)=3966 \begin{array}{l} \displaystyle (\sqrt{3}+\sqrt{2})^6-(\sqrt{3}-\sqrt{2})^6 \\[0.4em] \displaystyle =2[6(\sqrt{3})^5\sqrt{2}\allowbreak+20(\sqrt{3})^3(\sqrt{2})^3\allowbreak+6\sqrt{3}(\sqrt{2})^5] \\[0.4em] \displaystyle =2[6\times9\sqrt{3}\times\sqrt{2}\allowbreak+20\times3\sqrt{3}\times2\sqrt{2}\allowbreak+6\sqrt{3}\times4\sqrt{2}] \\[0.4em] \displaystyle =2(54\sqrt{6}+120\sqrt{6}+24\sqrt{6}) \\[0.4em] \displaystyle =2(198\sqrt{6}) \\[0.4em] \displaystyle =396\sqrt{6} \end{array}

Q3. Find 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.

Sol. The given expression is of the form (x+y)4+(x−y)4(x+y)^4+(x-y)^4.

By the binomial theorem,

(x+y)4=4C0x4+4C1x3y+4C2x2y2+4C3xy3+4C4y4(x−y)4=4C0x4−4C1x3y+4C2x2y2−4C3xy3+4C4y4 \begin{array}{l} \displaystyle (x+y)^4={}^4C_0x^4+{}^4C_1x^3y+{}^4C_2x^2y^2+{}^4C_3xy^3+{}^4C_4y^4 \\[0.4em] \displaystyle (x-y)^4={}^4C_0x^4-{}^4C_1x^3y+{}^4C_2x^2y^2-{}^4C_3xy^3+{}^4C_4y^4 \end{array}

Adding,

(x+y)4+(x−y)4=(4C0x4+4C1x3y+4C2x2y2+4C3xy3+4C4y4)+(4C0x4−4C1x3y+4C2x2y2−4C3xy3+4C4y4)=2(4C0x4+4C2x2y2+4C4y4)=2(x4+6x2y2+y4) \begin{array}{l} \displaystyle (x+y)^4+(x-y)^4 \\[0.4em] \displaystyle =({}^4C_0x^4+{}^4C_1x^3y+{}^4C_2x^2y^2\allowbreak+{}^4C_3xy^3+{}^4C_4y^4) \\[0.4em] \displaystyle \mathrel{+}({}^4C_0x^4-{}^4C_1x^3y+{}^4C_2x^2y^2\allowbreak-{}^4C_3xy^3+{}^4C_4y^4) \\[0.4em] \displaystyle =2({}^4C_0x^4+{}^4C_2x^2y^2+{}^4C_4y^4) \\[0.4em] \displaystyle =2(x^4+6x^2y^2+y^4) \end{array}

Here, x=a2x=a^2 and y=a2−1y=\sqrt{a^2-1}.

Substituting the values,

(a2+a2−1)4+(a2−a2−1)4=2[(a2)4+6(a2)2(a2−1)2+(a2−1)4]=2[a8+6a4(a2−1)+(a2−1)2]=2[a8+6a6−6a4+a4−2a2+1]=2[a8+6a6−5a4−2a2+1]=2a8+12a6−10a4−4a2+2 \begin{array}{l} \displaystyle \left(a^2+\sqrt{a^2-1}\right)^4+\left(a^2-\sqrt{a^2-1}\right)^4 \\[0.4em] \displaystyle =2\bigl[(a^2)^4\allowbreak+6(a^2)^2\left(\sqrt{a^2-1}\right)^2\allowbreak+\left(\sqrt{a^2-1}\right)^4\bigr] \\[0.4em] \displaystyle =2[a^8+6a^4(a^2-1)+(a^2-1)^2] \\[0.4em] \displaystyle =2[a^8+6a^6-6a^4+a^4-2a^2+1] \\[0.4em] \displaystyle =2[a^8+6a^6-5a^4-2a^2+1] \\[0.4em] \displaystyle =2a^8+12a^6-10a^4-4a^2+2 \end{array}

Q4. Find an approximation of (0.99)5(0.99)^5 using the first three terms of its expansion.

Sol. We have,

(0.99)5=(1−0.01)5 (0.99)^5=(1-0.01)^5

Here, a=1a=1, b=−0.01b=-0.01 and n=5n=5.

By the binomial theorem, taking the first three terms,

(a+b)n≈nC0an+nC1an−1b+nC2an−2b2\textcolor{#1B5E20}{ (a+b)^n\approx{}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2 }

Substituting the values,

(1−0.01)5=5C0(1)5+5C1(1)4(−0.01)+5C2(1)3(−0.01)2(0.99)5≈1×1−5×0.01+10×0.01×0.01=1−0.05+0.001=1.001−0.05=0.951 \begin{array}{l} \displaystyle (1-0.01)^5={}^5C_0(1)^5+{}^5C_1(1)^4(-0.01)+{}^5C_2(1)^3(-0.01)^2 \\[0.4em] \displaystyle (0.99)^5\approx1\times1-5\times0.01+10\times0.01\times0.01 \\[0.4em] \displaystyle =1-0.05+0.001 \\[0.4em] \displaystyle =1.001-0.05 \\[0.4em] \displaystyle =0.951 \end{array}

Q5. Expand using Binomial Theorem (1+x2−2x)4\left(1+\frac{x}{2}-\frac{2}{x}\right)^4, x≠0x\ne0.

Sol. Let y=1+x2y=1+\frac{x}{2}. Then

(1+x2−2x)4=[y+(−2x)]4 \left(1+\frac{x}{2}-\frac{2}{x}\right)^4=\left[y+\left(-\frac{2}{x}\right)\right]^4

By the binomial theorem,

(a+b)n=nC0an+nC1an−1b+nC2an−2b2+⋯+nCnbn\textcolor{#1B5E20}{ (a+b)^n={}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+\dots+{}^nC_nb^n }

Here, a=ya=y, b=−2xb=-\frac{2}{x} and n=4n=4.

Substituting the values,

[y+(−2x)]4=4C0y4+4C1y3(−2x)+4C2y2(−2x)2+4C3y(−2x)3+4C4(−2x)4 \begin{array}{l} \displaystyle \left[y+\left(-\frac{2}{x}\right)\right]^4\mathrel{=}{}^4C_0y^4+{}^4C_1y^3\left(-\frac{2}{x}\right)+{}^4C_2y^2\left(-\frac{2}{x}\right)^2+{}^4C_3y\left(-\frac{2}{x}\right)^3+{}^4C_4\left(-\frac{2}{x}\right)^4 \end{array}
=y4+4y3(−2x)+6y2(4x2)+4y(−8x3)+1(16x4)=y4−8xy3+24x2y2−32x3y+16x4...(1) \begin{array}{l} \displaystyle =y^4+4y^3\left(-\frac{2}{x}\right)+6y^2\left(\frac{4}{x^2}\right)+4y\left(-\frac{8}{x^3}\right)+1\left(\frac{16}{x^4}\right) \\[0.4em] \displaystyle =y^4-\frac{8}{x}y^3+\frac{24}{x^2}y^2-\frac{32}{x^3}y+\frac{16}{x^4}\qquad ...(1) \end{array}

Expanding the required powers of y by the binomial theorem:

Since y4=(1+x2)4y^4=\left(1+\frac{x}{2}\right)^4,

y4=4C0+4C1(x2)+4C2(x2)2+4C3(x2)3+4C4(x2)4=1+2x+3x22+x32+x416 \begin{array}{l} \displaystyle y^4={}^4C_0+{}^4C_1\left(\frac{x}{2}\right)+{}^4C_2\left(\frac{x}{2}\right)^2+{}^4C_3\left(\frac{x}{2}\right)^3+{}^4C_4\left(\frac{x}{2}\right)^4 \\[0.8em] \displaystyle =1+2x+\frac{3x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16} \end{array}

Since y3=(1+x2)3y^3=\left(1+\frac{x}{2}\right)^3,

y3=3C0+3C1(x2)+3C2(x2)2+3C3(x2)3=1+3x2+3x24+x38 \begin{array}{l} \displaystyle y^3={}^3C_0+{}^3C_1\left(\frac{x}{2}\right)+{}^3C_2\left(\frac{x}{2}\right)^2+{}^3C_3\left(\frac{x}{2}\right)^3 \\[0.8em] \displaystyle =1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8} \end{array}

Similarly, y2=(1+x2)2=1+x+x24y^2=\left(1+\frac{x}{2}\right)^2=1+x+\frac{x^2}{4}.

Substituting these in (1),

(1+x2−2x)4=(1+2x+3x22+x32+x416)−8x(1+3x2+3x24+x38)+24x2(1+x+x24)−32x3(1+x2)+16x4=1+2x+3x22+x32+x416−8x−12−6x−x2+24x2+24x+6−32x3−16x2+16x4 \begin{array}{l} \displaystyle \left(1+\frac{x}{2}-\frac{2}{x}\right)^4 \\[0.8em] \displaystyle =\left(1+2x+\frac{3x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}\right)-\frac{8}{x}\left(1+\frac{3x}{2}+\frac{3x^2}{4}+\frac{x^3}{8}\right)+\frac{24}{x^2}\left(1+x+\frac{x^2}{4}\right)-\frac{32}{x^3}\left(1+\frac{x}{2}\right)+\frac{16}{x^4} \\[0.8em] \displaystyle =1+2x+\frac{3x^2}{2}+\frac{x^3}{2}+\frac{x^4}{16}-\frac{8}{x}-12-6x-x^2+\frac{24}{x^2}+\frac{24}{x}+6-\frac{32}{x^3}-\frac{16}{x^2}+\frac{16}{x^4} \end{array}

Collecting like terms,

=x416+x32+(32−1)x2+(2−6)x+(1−12+6)+24−8x+24−16x2−32x3+16x4=x416+x32+x22−4x−5+16x+8x2−32x3+16x4 \begin{array}{l} \displaystyle =\frac{x^4}{16}+\frac{x^3}{2}+\left(\frac{3}{2}-1\right)x^2+(2-6)x+(1-12+6)+\frac{24-8}{x}+\frac{24-16}{x^2}-\frac{32}{x^3}+\frac{16}{x^4} \\[0.8em] \displaystyle =\frac{x^4}{16}+\frac{x^3}{2}+\frac{x^2}{2}-4x-5+\frac{16}{x}+\frac{8}{x^2}-\frac{32}{x^3}+\frac{16}{x^4} \end{array}

Q6. Find the expansion of (3x2−2ax+3a2)3(3x^2-2ax+3a^2)^3 using binomial theorem.

Sol. Let y=3x2−2axy=3x^2-2ax. Then (3x2−2ax+3a2)3=(y+3a2)3(3x^2-2ax+3a^2)^3=(y+3a^2)^3.

By the binomial theorem,

(p+q)n=nC0pn+nC1pn−1q+nC2pn−2q2+⋯+nCnqn\textcolor{#1B5E20}{ (p+q)^n={}^nC_0p^n+{}^nC_1p^{n-1}q+{}^nC_2p^{n-2}q^2+\dots+{}^nC_nq^n }

Here, p=yp=y, q=3a2q=3a^2 and n=3n=3.

Substituting the values,

(y+3a2)3=3C0y3+3C1y2(3a2)+3C2y(3a2)2+3C3(3a2)3=y3+9a2y2+27a4y+27a6...(1) \begin{array}{l} \displaystyle (y+3a^2)^3={}^3C_0y^3+{}^3C_1y^2(3a^2)+{}^3C_2y(3a^2)^2+{}^3C_3(3a^2)^3 \\[0.4em] \displaystyle =y^3+9a^2y^2+27a^4y+27a^6\qquad ...(1) \end{array}

Expanding the required powers of y by the binomial theorem:

Since y3=[3x2+(−2ax)]3y^3=[3x^2+(-2ax)]^3,

y3=3C0(3x2)3+3C1(3x2)2(−2ax)+3C2(3x2)(−2ax)2+3C3(−2ax)3=27x6+3(9x4)(−2ax)+3(3x2)(4a2x2)+1(−8a3x3)=27x6−54ax5+36a2x4−8a3x3 \begin{array}{l} \displaystyle y^3={}^3C_0(3x^2)^3+{}^3C_1(3x^2)^2(-2ax)+{}^3C_2(3x^2)(-2ax)^2+{}^3C_3(-2ax)^3 \\[0.4em] \displaystyle =27x^6+3(9x^4)(-2ax)+3(3x^2)(4a^2x^2)+1(-8a^3x^3) \\[0.4em] \displaystyle =27x^6-54ax^5+36a^2x^4-8a^3x^3 \end{array}

Since y2=[3x2+(−2ax)]2y^2=[3x^2+(-2ax)]^2,

y2=2C0(3x2)2+2C1(3x2)(−2ax)+2C2(−2ax)2=9x4−12ax3+4a2x2 \begin{array}{l} \displaystyle y^2={}^2C_0(3x^2)^2+{}^2C_1(3x^2)(-2ax)+{}^2C_2(-2ax)^2 \\[0.4em] \displaystyle =9x^4-12ax^3+4a^2x^2 \end{array}

Substituting these and y=3x2−2axy=3x^2-2ax in (1),

(3x2−2ax+3a2)3=(27x6−54ax5+36a2x4−8a3x3)+9a2(9x4−12ax3+4a2x2)+27a4(3x2−2ax)+27a6 \begin{array}{l} \displaystyle (3x^2-2ax+3a^2)^3\mathrel{=}(27x^6-54ax^5+36a^2x^4-8a^3x^3)+9a^2(9x^4-12ax^3+4a^2x^2)+27a^4(3x^2-2ax)+27a^6 \end{array}
=27x6−54ax5+36a2x4−8a3x3+81a2x4−108a3x3+36a4x2+81a4x2−54a5x+27a6 \begin{array}{l} \displaystyle =27x^6-54ax^5+36a^2x^4-8a^3x^3+81a^2x^4-108a^3x^3+36a^4x^2+81a^4x^2-54a^5x+27a^6 \end{array}

Collecting like terms,

=27x6−54ax5+117a2x4−116a3x3+117a4x2−54a5x+27a6 =27x^6-54ax^5+117a^2x^4-116a^3x^3+117a^4x^2-54a^5x+27a^6

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