Question

If ${\mathbf{C}}_{\mathbf{0}}\mathbf{,}{\mathbf{C}}_{\mathbf{1}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathbf{C}}_{\mathbf{\infty }}$ denote the binomial coefficients in the expansion of${\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}}^{\mathbf{n}}$. Then, the value of ${\mathbf{C}}_{\mathbf{1}}\mathbf{-}{}^{\mathbf{2}}\mathbf{C}_{\mathbf{2}}\mathbf{+}{}^{\mathbf{3}}\mathbf{C}_{\mathbf{3}}\mathbf{-}{}^{\mathbf{4}}\mathbf{C}_{\mathbf{4}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}$(up to n terms) is

• A. ${\mathbf{2}}^{\mathbf{n}}$
• B. ${\mathbf{2}}^{\mathbf{-}\mathbf{n}}$
• C. $\mathbf{0}$
• D. $\mathbf{1}$

Answer 1

The correct option is C

$\mathbf{0}$

Explanation for the correct options:

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

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

Step2 : Differentiating both sides with respect to $\mathit{x}$, we get

$n{(1+x)}^{\mathrm{x}-1}={\mathrm{C}}_1+2{\mathrm{C}}_2\mathrm{x}+3{\mathrm{C}}_3{\mathrm{x}}^2+....+{\mathrm{nC}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}$

putting $\mathit{x}\mathbf{=}\mathbf{-}\mathbf{1}$

then,

${\mathrm{C}}_1-2{\mathrm{C}}_2+3{\mathrm{C}}_3-....+{(-1)}^{\mathrm{n}-1}{\mathrm{nC}}_{\mathrm{n}}=0$

Hence, the correct option is (C)