Question

The sum of the last eight coefficients in the expansion of ${(1+x)}^{15}$ is

• A. $2^{16}$
• B. $2^{15}$
• C. $2^{14}$
• D. None of these

Answer 1

The correct option is C

$2^{14}$

Calculate the sum of the last eight coefficients in the given expansion

Given, ${(1+x)}^{15}$

From the binomial theorem, ${\left(a+b\right)}^n={}^n{C}_0{a}^n{b}^0+{}^n{C}_1{a}^{n-1}{b}^1+{}^n{C}_2{a}^{n-2}{b}^2+\dots +{}^n{C}_{n-1}{a}^1{b}^{n-1}+{}^n{C}_n{a}^0{b}^n$

Thus, the expansion of ${(1+x)}^{15}={}^{15}C_0·1^{15}·x^0+{}^{15}C_1·1^{14}{x}^1+{}^{15}C_2·1^{13}·x^2+{}^{15}C_3·1^{12}·x^3+\dots +{}^{15}C_{15}·1^0{x}^{15}$

The last $8$ coefficients are ${}^{15}C_8,{}^{15}C_9,\dots ,{}^{15}C_{15}$

Setting $x=1$ in the given expression turns the terms in the $RHS$ to have only the coefficients of the expansion.

$\begin{array}{ccc}\Rightarrow & {}^{15}C_0+{}^{15}C_1+\dots {}^{15}C_6+{}^{15}C_7+{}^{15}C_8+{}^{15}C_9\dots +{}^{15}C_{15}=2^{15}& \\ \Rightarrow & {}^{15}C_{15}+{}^{15}C_{14}+\dots +{}^{15}C_9+{}^{15}C_8+{}^{15}C_8+{}^{15}C_9+\dots +{}^{15}C_{15}=2^{15}& \left[\because {}^{n}C_r={}^{n}C_{n-r}\right]\\ \Rightarrow & 2\left({}^{15}C_8+{}^{15}C_9+\dots +{}^{15}C_{15}\right)=2^{15}& \\ \Rightarrow & {}^{15}C_8+{}^{15}C_9+\dots +{}^{15}C_{15}=2^{14}& \end{array}$

Hence, option (C) is correct.