Question

The value of $-{}^{15}C_1+2·{}^{15}C_2-3·{}^{15}C_3+\dots -15·{}^{15}C_{15}+{}^{14}C_1+{}^{14}C_3+{}^{14}C_5+\dots +{}^{14}C_{11}$ is

• A. $2^{14}$
• B. $2^{13}-13$
• C. $2^{16}-1$
• D. $2^{13}-14$

Answer 1

The correct option is D

$2^{13}-14$

Explanation for the correct option:

Given, $-{}^{15}C_1+2·{}^{15}C_2-3·{}^{15}C_3+\dots -15·{}^{15}C_{15}+{}^{14}C_1+{}^{14}C_3+{}^{14}C_5+\dots +{}^{14}C_{11}$

Let,
$\begin{array}{rcl}{S}_1& =& -{}^{15}C_1+2·{}^{15}C_2-3·{}^{15}C_3+\dots -15·{}^{15}C_{15}\\ & =& \sum _{r=1}^{15}{\left(-1\right)}^r·r·{}^{15}C_r\\ & =& 15·\sum _{r=1}^{15}\left[{\left(-1\right)}^r{}^{14}C_{r-1}\right]\mathbf{}\mathbf{[}\mathbf{\because }{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{r}}{}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{C}_{\mathbf{r}\mathbf{-}\mathbf{1}}\mathbf{]}\\ & =& 15\left(-{}^{14}C_0+{}^{14}C_1-{}^{14}C_2+\dots -{}^{14}C_{14}\right)\left[{\mathbf{(}\mathbf{1}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{14}}\mathbf{=}{}^{\mathbf{14}}\mathbf{C}_{\mathbf{0}}\mathbf{-}{}^{\mathbf{14}}\mathbf{C}_{\mathbf{1}}\mathbf{x}\mathbf{+}{}^{\mathbf{14}}\mathbf{C}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{\dots }\mathbf{+}{}^{\mathbf{14}}\mathbf{C}_{\mathbf{14}}{\mathbf{x}}^{\mathbf{14}}\right]\\ & =& 15\left(0\right)\\ & =& 0\end{array}$

Let,
$\begin{array}{rcl}{S}_2& =& {}^{14}C_1+{}^{14}C_3+{}^{14}C_5+\dots +{}^{14}C_{11}\\ & =& \left({}^{14}C_1+{}^{14}C_3+{}^{14}C_5+\dots +{}^{14}C_{11}+{}^{14}C_{13}\right)-{}^{14}C_{13}\\ & =& 2^{14-1}-{}^{14}C_{13}\mathbf{}\left[\because {}^{n}C_1+{}^{n}C_3+{}^{n}C_5+.....=2^{n-1}\right]\\ & =& 2^{13}-14\end{array}$

So, $S_1+S_2=0+2^{13}-14=2^{13}-14$

Therefore, $-{}^{15}C_1+2·{}^{15}C_2-3·{}^{15}C_3+\dots -15·{}^{15}C_{15}+{}^{14}C_1+{}^{14}C_3+{}^{14}C_5+\dots +{}^{14}C_{11}=2^{13}-14$

Hence, option D is correct.