Question

The value of $1^2.{\mathrm{C}}_1+3^2.{\mathrm{C}}_3+5^2.{\mathrm{C}}_5+...$ is

• A. $\mathrm{n}{(\mathrm{n}–1)}^{\mathrm{n}–2}+\mathrm{n}{2}^{\mathrm{n}–1}$
• B. $\mathrm{n}(\mathrm{n}–1)2^{\mathrm{n}-2}$
• C. $\mathrm{n}(\mathrm{n}–1)2^{\mathrm{n}-3}$
• D. None of these

Answer 1

The correct option is D

None of these

The explanation for the correct answer.

Step-1 Summation of binomial coefficient series using differentiation:

Given: $1^2.{\mathrm{C}}_1+3^2.{\mathrm{C}}_3+5^2.{\mathrm{C}}_5+...$

Binomial expansion of ${\left(1+x\right)}^{\mathbf{n}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{C}}_{\mathbf{0}}\mathbf{+}{\mathit{C}}_{\mathbf{1}}\mathit{x}\mathbf{+}{\mathit{C}}_{\mathbf{2}}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}}$

Differentiate the expansion with respect to $\mathrm{x}$

$\Rightarrow \mathrm{n}{\left(1+\mathrm{x}\right)}^{\mathrm{n}-1}={\mathrm{C}}_1+2{\mathrm{C}}_2\mathrm{x}+.......+{\mathrm{nC}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}-1}\mathbf{}\left[\frac{d}{\mathrm{dx}}{\left(x\right)}^n={\mathrm{nx}}^{n-1}\right]$

Multiply $\mathrm{x}$ on both sides of the above equation.

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

Differentiate the equation again with respect to $\mathrm{x}$

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

Step-2: Summation of series:

Put $\mathrm{x}=1$in equation (1)

$$\begin{gathered} \Rightarrow \mathrm{n}{\left(1+1\right)}^{\mathrm{n}-1}+\mathrm{n}1(\mathrm{n}-1){(1+1)}^{\mathrm{n}-2}={\mathrm{C}}_1+2^2{\mathrm{C}}_{2}1+.......+{\mathrm{n}}^2{\mathrm{C}}_{\mathrm{n}}{1}^{\mathrm{n}-1} \\ \Rightarrow \mathrm{n}{2}^{\mathrm{n}-1}+\mathrm{n}(\mathrm{n}-1)2^{\mathrm{n}-2}={\mathrm{C}}_1+2^2{\mathrm{C}}_2+.......+{\mathrm{n}}^2{\mathrm{C}}_{\mathrm{n}}-\left(2\right) \end{gathered}$$

Put the value of $\mathrm{x}=-1$in equation (1)

$$\begin{gathered} \Rightarrow \mathrm{n}{\left(1-1\right)}^{\mathrm{n}-1}+\mathrm{n}(-1)(\mathrm{n}-1){(1-1)}^{\mathrm{n}-2}={\mathrm{C}}_1+2^2{\mathrm{C}}_2(-1)+.......+{\mathrm{n}}^2{\mathrm{C}}_{\mathrm{n}}{(-1)}^{\mathrm{n}-1} \\ \Rightarrow 0={\mathrm{C}}_1-2^2{\mathrm{C}}_2+.......-\left(3\right) \end{gathered}$$

Add equations (2) and (3)

$$\begin{gathered} \Rightarrow \mathrm{n}{2}^{\mathrm{n}-1}+\mathrm{n}(\mathrm{n}-1)2^{\mathrm{n}-2}=2\left({\mathrm{C}}_1+3^2{\mathrm{C}}_3+5^2{\mathrm{C}}_{5+........}\right) \\ \Rightarrow \frac{\mathrm{n}{2}^{\mathrm{n}-1}+\mathrm{n}(\mathrm{n}-1)2^{\mathrm{n}-2}}{2}={\mathrm{C}}_1+3^2{\mathrm{C}}_3+5^2{\mathrm{C}}_{5+........} \end{gathered}$$

Left-hand side of the equation is

$$\begin{gathered} =\frac{2^{\mathrm{n}-2}\mathrm{n}\left(2+(\mathrm{n}-1)\right)}{2} \\ =2^{\mathrm{n}-3}\mathrm{n}\left(\mathrm{n}+1\right) \end{gathered}$$

Hence,option() is the correct answer.