Question

In Problems 25-30 proceed as in Example 3 to rewrite the given expression using a single power series whose general term involves ${\mathit{x}}^{\mathbf{k}}$.

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathit{n}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{3}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}\mathbf{+}\mathbf{2}}$

Short answer

$\sum _{n=1}^{\infty }n{C}_n{x}^{n-1}+3\sum _{n=0}^{\infty }{C}_n{x}^{n+2}=C_1+2C_{2}x+\sum _{k=2}^{\infty }\left[(k+1)C_{k+1}+3C_{k-2}\right]x^k$

Step-by-step solution

Consider the power series

The power series is,

$\sum _{n=1}^{\infty }n{C}_n{x}^{n-1}+3\sum _{n=0}^{\infty }{C}_n{x}^{n+2}=C_1+2C_{2}x+\sum _{n=3}^{\infty }n{C}_n{x}^{n-1}+3\sum _{n=0}^{\infty }{C}_n{x}^{n+2}$

Compute the limits

Let$n=k+1$for first series

This implies,$k=n-1$

The limits of the series will be,

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

Upper limit,$n=\infty \Rightarrow k=\infty -1=\infty$

Let$n=k-2$for second series

This implies,$k=n+2$

The limits of the series will be,

Lower limit,$n=0\Rightarrow k=0+2=2$

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

Use the substitute

Substitute the value of $k$in the series.

$$\begin{gathered} C_1+2C_{2}x+\sum _{n=3}^{\infty }n{C}_n{x}^{n-1}+3\sum _{n=0}^{\infty }{C}_n{x}^{n+2}=C_1+2C_{2}x+\sum _{k=2}^{\infty }(k+1)C_{k+1}{x}^k+3\sum _{k=2}^{\infty }{C}_{k-2}{x}^k \\ =C_1+2C_{2}x+\sum _{k=2}^{\infty }\left[(k+1)C_{k+1}+3C_{k-2}\right]x^k \end{gathered}$$

Therefore, the power series is,

$\sum _{n=1}^{\infty }n{C}_n{x}^{n-1}+3\sum _{n=0}^{\infty }{C}_n{x}^{n+2}=C_1+2C_{2}x+\sum _{k=2}^{\infty }\left[(k+1)C_{k+1}+3C_{k-2}\right]x^k$