Question

Verify the given equation. $$ \sum_{n=0}^{\infty} a_{n}(x-1)^{n+1}=\sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$

Short answer

Question: Verify that the following equation is true: $$ \sum_{n=0}^{\infty} a_{n}(x-1)^{n+1}=\sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$ Answer: We can verify the given equation by changing the indices of the left side series to start at \(n=1\), simplifying the series terms, and showing that the terms of both series are equal. Therefore, we can conclude that the equation is true: $$ \sum_{m=0}^{\infty} a_{m+1}(x-1)^{m+2} = \sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$

Step-by-step solution

Analyze the given equation

First, let us analyze the equation. $$ \sum_{n=0}^{\infty} a_{n}(x-1)^{n+1}=\sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$ We see that on the left side, the series starts at \(n = 0\) and on the right side, the series starts at \(n = 1\). Our goal is to change the left side series to start at \(n=1\) and adjust the terms accordingly.

Change the indices of the series

To change the left side series summation to start from \(n=1\), we need to replace variable \(n\) with a new variable, for example \(m = n-1\). Then, when \(n=0 \Rightarrow m=-1\). Therefore, when \(n=1\), so \(m=0\). Moreover, when \(m = n-1\), then when \(m+1=n\). So we have: $$ \sum_{m=1-1}^{\infty} a_{m+1}(x-1)^{(m+1)+1}=\sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$ Now, our left side starts at \(n=1\), so we can rewrite our equation as: $$ \sum_{m=0}^{\infty} a_{m+1}(x-1)^{(m+1)+1}=\sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$

Simplify the series terms

Now, let's simplify the left side series terms. $$ \sum_{m=0}^{\infty} a_{m+1}(x-1)^{(m+1)+1} = \sum_{m=0}^{\infty} a_{m+1}(x-1)^{m+2} $$

Compare the series

Now that we have both series in the same form, we can compare their terms. Notice that for every natural number \(m\) in the left side series, \(a_{m+1}(x-1)^{m+2}\), the terms are equal to the terms at the right side series, \(a_{n-1}(x-1)^{n}\) , for every natural number \(n\). We've verified our given equation: $$ \sum_{m=0}^{\infty} a_{m+1}(x-1)^{m+2} = \sum_{n=1}^{\infty} a_{n-1}(x-1)^{n} $$