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Binomial Theorem

Class 11 · Maths · Chapter 7 · Examples

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.

Examples

Eg 1. Expand (x2+3x)4\left(x^2+\frac{3}{x}\right)^4, x≠0x\ne0

Sol. 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=x2a=x^2, b=3xb=\frac{3}{x} and n=4n=4.

Substituting the values,

(x2+3x)4=4C0(x2)4+4C1(x2)3(3x)+4C2(x2)2(3x)2+4C3(x2)(3x)3+4C4(3x)4 \begin{array}{l} \displaystyle \left(x^2+\frac{3}{x}\right)^4\mathrel{=}{}^4C_0(x^2)^4+{}^4C_1(x^2)^3\left(\frac{3}{x}\right)+{}^4C_2(x^2)^2\left(\frac{3}{x}\right)^2+{}^4C_3(x^2)\left(\frac{3}{x}\right)^3+{}^4C_4\left(\frac{3}{x}\right)^4 \end{array}
=1(x8)+4x6(3x)+6x4(9x2)+4x2(27x3)+1(81x4)=x8+12x5+54x2+108x+81x4 \begin{array}{l} \displaystyle =1(x^8)+4x^6\left(\frac{3}{x}\right)+6x^4\left(\frac{9}{x^2}\right)+4x^2\left(\frac{27}{x^3}\right)+1\left(\frac{81}{x^4}\right) \\[0.4em] \displaystyle =x^8+12x^5+54x^2+\frac{108}{x}+\frac{81}{x^4} \end{array}

Eg 2. Compute (98)5(98)^5.

Sol. We write 98=100+(−2)98=100+(-2), so (98)5=[100+(−2)]5(98)^5=[100+(-2)]^5.

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=100a=100, b=−2b=-2 and n=5n=5.

Substituting the values,

(98)5=5C0(100)5+5C1(100)4(−2)+5C2(100)3(−2)2+5C3(100)2(−2)3+5C4(100)(−2)4+5C5(−2)5 \begin{array}{l} \displaystyle (98)^5\mathrel{=}{}^5C_0(100)^5+{}^5C_1(100)^4(-2)+{}^5C_2(100)^3(-2)^2+{}^5C_3(100)^2(-2)^3+{}^5C_4(100)(-2)^4+{}^5C_5(-2)^5 \end{array}
=1×1010+5×108×(−2)+10×106×4+10×104×(−8)+5×100×16+1×(−32)=10000000000−1000000000+40000000−800000+8000−32=(10000000000+40000000+8000)−(1000000000+800000+32)=10040008000−1000800032=9039207968 \begin{array}{l} \displaystyle =1\times10^{10}+5\times10^8\times(-2)\allowbreak+10\times10^6\times4+10\times10^4\times(-8)\allowbreak+5\times100\times16+1\times(-32) \\[0.4em] \displaystyle =10000000000-1000000000+40000000-800000+8000-32 \\[0.4em] \displaystyle =(10000000000+40000000+8000)-(1000000000+800000+32) \\[0.4em] \displaystyle =10040008000-1000800032 \\[0.4em] \displaystyle =9039207968 \end{array}

Eg 3. Which is larger (1.01)1000000(1.01)^{1000000} or 10,000?

Sol. We write 1.01=1+0.011.01=1+0.01, so (1.01)1000000=(1+0.01)1000000(1.01)^{1000000}=(1+0.01)^{1000000}.

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=1a=1, b=0.01b=0.01 and n=1000000n=1000000.

Since b>0b>0, every term of the expansion is positive.

Substituting the values,

(1.01)1000000=1000000C0(1)1000000+1000000C1(1)999999(0.01)+other positive terms=1+1000000×0.01+other positive terms=1+10000+other positive terms=10001+other positive terms>10000 \begin{array}{l} \displaystyle (1.01)^{1000000}={}^{1000000}C_0(1)^{1000000}\allowbreak+{}^{1000000}C_1(1)^{999999}(0.01)\allowbreak+\text{other positive terms} \\[0.4em] \displaystyle =1+1000000\times0.01+\text{other positive terms} \\[0.4em] \displaystyle =1+10000+\text{other positive terms} \\[0.4em] \displaystyle =10001+\text{other positive terms}>10000 \end{array}

Since all the remaining terms are positive,

Hence, (1.01)1000000(1.01)^{1000000} is the larger number.

Eg 4. Using binomial theorem, prove that 6n−5n6^n-5n always leaves remainder 1 when divided by 25.

Sol. To show that 6n−5n6^n-5n leaves remainder 1 when divided by 25, we have to show that

6n−5n=25k+1 6^n-5n=25k+1

where kk is a whole number.

We write 6=1+56=1+5, so

6n=(1+5)n 6^n=(1+5)^n

By the binomial theorem,

(a+b)n=nC0an+nC1an−1b+nC2an−2b2+nC3an−3b3+⋯+nCnbn (a+b)^n={}^nC_0a^n+{}^nC_1a^{n-1}b+{}^nC_2a^{n-2}b^2+{}^nC_3a^{n-3}b^3+\dots+{}^nC_nb^n

Here, a=1a=1 and b=5b=5.

Substituting the values,

(1+5)n=nC0(1)n+nC1(1)n−1(5)+nC2(1)n−2(5)2+nC3(1)n−3(5)3+⋯+nCn(5)n6n=1+5n+52 nC2+53 nC3+⋯+5n6n−5n=1+52 nC2+53 nC3+⋯+5n \begin{array}{l} \displaystyle (1+5)^n={}^nC_0(1)^n+{}^nC_1(1)^{n-1}(5)+{}^nC_2(1)^{n-2}(5)^2+{}^nC_3(1)^{n-3}(5)^3+\dots+{}^nC_n(5)^n \\[0.4em] \displaystyle 6^n=1+5n+5^2\,{}^nC_2+5^3\,{}^nC_3+\dots+5^n \\[0.4em] \displaystyle 6^n-5n=1+5^2\,{}^nC_2+5^3\,{}^nC_3+\dots+5^n \end{array}

Taking 525^2 common,

6n−5n=1+52[nC2+5 nC3+⋯+5n−2]=1+25k=25k+1 \begin{array}{l} \displaystyle 6^n-5n=1+5^2[{}^nC_2+5\,{}^nC_3+\dots+5^{n-2}] \\[0.4em] \displaystyle =1+25k \\[0.4em] \displaystyle =25k+1 \end{array}

where kk is a whole number.

For n=1n=1,

61−5(1)=1=25(0)+1 6^1-5(1)=1=25(0)+1

Since 0≤1<250\leq1<25, the remainder is 1.

∴ 6n−5n6^n-5n always leaves remainder 1 when divided by 25.

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