Index of Summation Shift? Power Series and Differential Equations I have never had to index shift a summation series before, and it seems relatively straightforward, however, I am looking at an example in my textbook that doesn't make sense.  I am wondering if someone might be able to outline the steps that appear to be missing.
Apologies for the formatting issue; perhaps someone can help with that?
Here is how it appears in the textbook:
$(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n$
$= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + \sum_{n=0}^{\infty}[2n(n-1)+6n+2]a_{n}x^n$
So it looks as though the two rightmost terms were combined somehow, but I cannot figure how this took place.
Can someone help with my understanding?  I've never done anything like this before, and I'm frustrated I can't understand the rest of the example because of this.
Thanks!
 A: Ok, just distribute to see:
\begin{align}
&(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n  =\\
&=\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +2x^2\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2}+
\sum_{n=1}^{\infty}6na_{n}xx^{n-1}+\sum_{n=0}^{\infty}2a_{n}x^n \\ 
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty}2n(n-1)a{_{n}}x^2x^{n-2}+
\sum_{n=1}^{\infty}6na_{n}x^{n}+\sum_{n=0}^{\infty}2a_{n}x^n \\ 
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty}2n(n-1)a{_{n}}x^{n}+
\sum_{n=2}^{\infty}6na_{n}x^{n}+\sum_{n=2}^{\infty}2a_{n}x^n +6a_1x+2a_0+2a_1x \\ 
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=2}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n +8a_1x+2a_0\\ 
\end{align}
Now, notice the rightmost sum begins at $n=0$ in your post, can we see the exceptional terms as the $n=0$ and $n=1$ terms in the text's expression?
Added: I thought it might be useful to show how to group all the terms: continuing,
\begin{align}
&(1+2x^2)\sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} + 6x\sum_{n=1}^{\infty}na_{n}x^{n-1}+2\sum_{n=0}^{\infty}a_{n}x^n  =\\
&= \sum_{n=2}^{\infty}n(n-1)a{_{n}}x^{n-2} +\sum_{n=0}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n \\ 
&= \sum_{j=0}^{\infty}(j+2)(j+1)a{_{j+2}}x^{j} +\sum_{n=0}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n \\ 
&= \sum_{j=0}^{\infty}(j+2)(j+1)a{_{j+2}}x^{j} +\sum_{n=0}^{\infty} \left( 2n(n-1)+
6n+2 \right)a_nx^n \\ 
&= \sum_{n=0}^{\infty} \left( [2n(n-1)+
6n+2]a_n+(n+2)(n+1)a{_{n+2}} \right)x^n 
\end{align}
So, then when we have this is zero we find relations between the $n$-th and $n+2$-th coefficients.
