Question

If ${\mathbf{C}}_{\mathbf{1}}\mathbf{,}{\mathbf{C}}_{\mathbf{2}}\mathbf{,}{\mathbf{C}}_{\mathbf{3}}\mathbf{}$are the usual binomial coefficients and $\mathbf{S}\mathbf{=}{\mathbf{C}}_{\mathbf{1}}\mathbf{+}\mathbf{2}{\mathbf{C}}_{\mathbf{2}}\mathbf{+}\mathbf{3}{\mathbf{C}}_{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathit{n}{\mathbf{C}}_{\mathbf{n}}$, then $\mathbf{S}$ equals

• A. $\mathit{n}{\mathbf{2}}^{\mathbf{n}}$
• B. ${\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$
• C. $\mathit{n}{\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$
• D. ${\mathbf{2}}^{\mathbf{n}\mathbf{+}\mathbf{1}}$

Answer 1

The correct option is C

$\mathit{n}{\mathbf{2}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$

Explanation for the correct options:

Step1 : Expansion of ${\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}}^{\mathbf{n}}$

${\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}}^{\mathbf{n}}\mathbf{=}{\mathbf{C}}_0\mathbf{+}{\mathbf{C}}_{\mathbf{1}}\mathbf{x}\mathbf{+}{\mathbf{C}}_{\mathbf{2}}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathbf{C}}_{\mathbf{n}}{\mathbf{x}}^{\mathbf{n}}$

Step2 : Differentiating with respect to x

$n{(1+x)}^n={\mathrm{C}}_1+2{\mathrm{C}}_2\mathrm{x}+...+{\mathrm{nC}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}$

Put $\mathit{x}\mathbf{=}\mathbf{1}$,

${\mathrm{C}}_1+{}^2\mathrm{C}_2+....+{}^{\mathrm{n}}\mathrm{C}_{\mathrm{n}}=\mathrm{n}{2}^{\mathrm{n}-1}$

Hence, the correct option is (C)