Question: Show that,

$X^2.\sum _{n-0}^{\infty }n(n+1)a_n{x}^n=\sum _{n-2}^{\infty }(n-2)(n-1)a_{n-2}{x}^n$

A power series is an infinite series of the form,$\sum _{n=0}^{\infty }{a}_n{(x-c)}^n=a_0+a_1(x-c)+a_2(x-c{)}^2+.....$

Where,a_n represents the coefficient term of the nth term and c is a constant.

We have to show that, .$X^2.\sum _{n-0}^{\infty }n(n+1)a_n{x}^n=\sum _{n-2}^{\infty }(n-2)(n-1)a_{n-2}{x}^n$

Simplifying the L.H.S expression,

$X^2.\sum _{n-0}^{\infty }n(n+1)a_n{x}^n=\sum _{n-0}^{\infty }n(n+1)a_{n-2}{x}^{n+2}$

Now changing the index, let,

$$\begin{gathered} n+2=k \\ n=k-2 \end{gathered}$$

Then,

role="math" localid="1664273606231" $\sum _{n-0}^{\infty }n(n+1)a_n{x}^{n-2}=\sum _{k-2-0}^{\infty }(k-2)(k-2+1)a_{k-2}{x}^k$

role="math" localid="1664273763077" $=\sum _{k-2}^{\infty }(k-2)(k-1)a_{k-2}{x}^k$

The index is a dummy variable, so we can replace k with n , the expressionbecomes,

role="math" localid="1664273826337" $\sum _{n-2}^{\infty }(n-2)(n-1)a_{n-2}{x}^n$which is equal to the R.H.S of the given statement

Hence proved$X^2.\sum _{n-0}^{\infty }n(n+1)a_n{x}^n=\sum _{n-2}^{\infty }(n-2)(n-1)a_{n-2}{x}^n$