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Expert-verified Found in: Page 1 ### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069 # Question: Show that,$28.\sum _{n=0}^{\infty }{a}_{n}{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_{n}{x}^{n-1}={b}_{1}+\sum _{n-1}^{\infty }\left[2{a}_{n-1}+\left(n+1\right){b}_{n+1}\right]{x}^{n}$

We showed that 2 .$\sum _{n=0}^{\infty }{a}_{n}{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_{n}{x}^{n-1}={b}_{1}+\sum _{n-1}^{\infty }\left[2{a}_{n-1}+\left(n+1\right){b}_{n+1}\right]{x}^{n}$

See the step by step solution

## Step 1: Power series

A power series is an infinite series of the form,$\sum _{n=0}^{\infty }{a}_{n}{\left(x-c\right)}^{n}={a}_{0}+{a}_{1}\left(x-c\right)+{a}_{2}\left(x-c{\right)}^{2}+.....$Where,an represents the coefficient term of the nth term and c is a constant.

## Step 2: Changing the index of the first term

We have to show that,

2.$\sum _{n=0}^{\infty }{a}_{n}{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_{n}{x}^{n-1}={b}_{1}+\sum _{n-1}^{\infty }\left[2{a}_{n-1}+\left(n+1\right){b}_{n+1}\right]{x}^{n}$

Simplifying the L.H.S expression,

L.H.S =2.$\sum _{n=0}^{\infty }{a}_{n}{x}^{n+1}+\sum _{n-1}^{\infty }n{b}_{n}{x}^{n-1}$

Now changing the index of the first term, let,

n + 1 = k

n = k -1

Then,

$2\sum _{n=0}^{\infty }{a}_{n}{x}^{n+1}=2\sum _{k-1=0}^{\infty }{a}_{k-1}{x}^{k}\phantom{\rule{0ex}{0ex}}=2\sum _{k=0}^{\infty }{a}_{k-1}{x}^{k}$

The index is a dummy variable, so we can replace k with n , the first term of the L.H.S becomes$,2\sum _{k=0}^{\infty }{a}_{k-1}{x}^{k}$

## Step 3: Changing the index of the second term

Now changing the index of the second term, let,

n - 1 = k

n = k + 1

Then,

$\sum _{n-1}^{\infty }n{b}_{n}{x}^{n-1}=\sum _{k+1=1}^{\infty }\left(k+1\right){b}_{k+1}{x}^{k}\phantom{\rule{0ex}{0ex}}=\sum _{k=0}^{\infty }\left(k+1\right){b}_{k+1}{x}^{k}$

The index is a dummy variable, so we can replace k with n , the first term of the L.H.S becomes $=\sum _{k=0}^{\infty }\left(k+1\right){b}_{k+1}{x}^{k}$

The second expression in L.H.S, starts with index 0 , in order to combine both the expressions, we will take the second expression from n=1

,$2.\sum _{n=0}^{\infty }{a}_{n}{x}^{n+1}+\sum _{n=1}^{\infty }n{b}_{n}{x}^{n-1}=2.\sum _{n=1}^{\infty }{a}_{n-1}{x}^{n}+{b}_{1}+\sum _{n=0}^{\infty }\left(n+1\right){b}_{n+1}{x}^{n}\phantom{\rule{0ex}{0ex}}={b}_{1}+\sum _{n=1}^{\infty }\left[2{a}_{n-1}+\left(n+1\right){b}_{n+1}\right]{x}^{n}$

which is equal to the R.H.S of the given statement.

Therefore, by simplifying the L.H.S of the expression, we can prove that both the expressions are the same. ### Want to see more solutions like these? 