Question

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

$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\mathbf{2}\mathit{n}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{0}}^{\mathbf{\infty }}\mathbf{6}{\mathit{C}}_{\mathbf{n}}{\mathit{x}}^{\mathbf{n}\mathbf{+}\mathbf{1}}$

Short answer

$\sum _{n=1}^{\infty }2n{C}_n{x}^{n-1}+\sum _{n=0}^{\infty }6C_n{x}^{n+1}=2C_1+\sum _{k=1}^{\infty }\left[2(k+1)C_{k+1}+6C_{k-1}\right]x^k$

Step-by-step solution

Consider the power series

The power series is,

$$\begin{gathered} \sum 2n{C}_n{x}^{n-1}+\sum _{n=0}^{\infty }6C_n{x}^{n+1} \\ =2C_1+\sum _{n=2}^{\infty }2n{C}_n{x}^{n-1}+\sum _{n=0}^{\infty }6C_n{x}^{n+1} \end{gathered}$$

Compute the limits

Let$n=k+1$for first series

This implies,role="math" localid="1663929970361" $k=n-1$

The limits of the series will be,

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

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

Let $n=k-1$ for second series

This implies,$k=n+1$

The limits of the series will be,

Lower limit,$n=0\Rightarrow k=0+1=1$

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

Use the substitute

Substitute the value of$k$in the series.

$$\begin{gathered} 2C_1+\sum _{n=2}^{\infty }2n{C}_n{x}^{n-1}+\sum _{n=0}^{\infty }6C_n{x}^{n+1}=2C_1+\sum _{k=1}^{\infty }2(k+1)C_{k+1}{x}^k+6\sum _{k=1}^{\infty }{C}_{k-1}{x}^k \\ =2C_1+\sum _{k=1}^{\infty }\left[2(k+1)C_{k+1}+6C_{k-1}\right]x^k \end{gathered}$$

Therefore, the power series is,

$\sum _{n=1}^{\infty }2n{C}_n{x}^{n-1}+\sum _{n=0}^{\infty }6C_n{x}^{n+1}=2C_1+\sum _{k=1}^{\infty }\left[2(k+1)C_{k+1}+6C_{k-1}\right]x^k$