Question

In Problems 23 and 24 use a substitution to shift the summation index so that the general term of given power series involves ${\mathit{x}}^{\mathbf{k}}$.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathit{n}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}\mathbf{+}\mathbf{2}}$

Short answer

$\sum _{n=1}^{\infty }n{C}_n{x}^{n+2}=\sum _{k=3}^{\infty }(k-2)C_{k-2}{x}^k$

Step-by-step solution

Consider the power series

The power series is,

$\sum _{n=1}^{\infty }n{C}_n{x}^{n+2}$

Compute the limits

Let$n=k-2$

This implies,$k=n+2$

The limits of the series will be,

Lower limit,$n=1\Rightarrow k=1+2=3$

Upper limit,$n=\infty \Rightarrow k=\infty +2=\infty$

Use the substitute

Substitute the value of$k$ in the series.

$\sum _{n=1}^{\infty }n{C}_n{x}^{n+2}=\sum _{k=3}^{\infty }(k-2)C_{k-2}{x}^k$

Therefore, the power series is,

$\sum _{n=1}^{\infty }n{C}_n{x}^{n+2}=\sum _{k=3}^{\infty }(k-2)C_{k-2}{x}^k$