Question

${}^{15}C_3+{}^{15}C_5+...+{}^{15}C_{15}$ is equal to

• A. $2^{14}$
• B. $2^{14}-15$
• C. $2^{14}+15$
• D. $2^{14}-1$

Answer 1

The correct option is B

$2^{14}-15$

Explanation for the correct option:

Step 1. Find the summation of given combinations:

As we know,

${\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathit{C}_{\mathbf{0}}\mathbf{+}{}^{\mathbf{n}}\mathit{C}_{\mathbf{1}}\mathit{x}\mathbf{+}{}^{\mathbf{n}}\mathit{C}_{\mathbf{3}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{}^{\mathbf{n}}\mathit{C}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}$

Put $n=15$ in above expansion, we get

${\left(1+x\right)}^{15}={}^{15}C_0+{}^{15}C_{1}x+{}^{15}C_2{x}^2+.....+{}^{15}C_{15}{x}^{15}$

Put $x=1$ and $x=-1$ in above expansion, we get

${\left(2\right)}^{15}={}^{15}C_0+{}^{15}C_1+{}^{15}C_2+.....+{}^{15}C_{15}$ ……….(1)

$0={}^{15}C_0-{}^{15}C_{1}x+{}^{15}C_2-{}^{15}C_3+.....-{}^{15}C_{15}{x}^{15}$ ………(2)

Step 2. subtract equation (2) from equation (1):

$\Rightarrow$ $2^{15}=2\left({}^{15}C_1+{}^{15}C_3+{}^{15}C_5+......+{}^{15}C_{15}\right)$

$\Rightarrow$ $2^{14}=\left(15+{}^{15}C_3+{}^{15}C_5+......+{}^{15}C_{15}\right)$

$\Rightarrow 2^{14}-15=\left({}^{15}C_3+{}^{15}C_5+{}^{15}C_7+......+{}^{15}C_{15}\right)$

Hence, Option ‘B’ is Correct.