Q- 27E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Question: Show that,

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

Step-by-Step Solution

Answer

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

Step 1: Power series.

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.

Step 2: To express the given series in terms of generic term x k .

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,

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

                                  $= \sum _{k-2}^{\infty }(k-2)(k-1)a_{k-2}{x}^k$

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

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

 

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