Simplify a summation $ (1 - 2xt + x^2) \sum_{n=0}^{\infty}n P_n(t) x^{n-1} + (x - t) \sum_{n=0}^{\infty} P_n(t)x^n = 0 $ 
How to simplify,
from
$$ (1 - 2xt + x^2) \sum_{n=0}^{\infty}n P_n(t) x^{n-1} + (x - t) \sum_{n=0}^{\infty} P_n(t)x^n = 0 $$
to
$$P_1(t) - t P_0(t) + \sum_{n=1}^{\infty}[ (n+1) P_{n+1}(t) - (2n+1)tP_n(t) + n P_{n-1}(t)] x^n = 0$$

I understood that I have to expand , collect the power, and shift some index.
Given:
$$ (1 - 2xt + x^2) \sum_{n=0}^{\infty}n P_n(t) x^{n-1} + (x - t) \sum_{n=0}^{\infty} P_n(t)x^n = 0 $$
Expanding
$$\sum_{n=0}^{\infty}n P_n(t) x^{n-1} -2t \sum_{n=0}^{\infty}n P_n(t) x^{n} + \sum_{n=0}^{\infty}n P_n(t) x^{n+1} + \sum_{n=0}^{\infty} P_n(t) x^{n+1}-  \sum_{n=0}^{\infty} P_n(t) x^{n}$$
I don't see how I could collect the power, and shifting the index is involved. I need help to understand this. Any input is much appreciated.
 A: Write $$\sum_{n=0}^{\infty}n P_n(t) x^{n-1} -2t \sum_{n=0}^{\infty}n P_n(t) x^{n} + \sum_{n=0}^{\infty}n P_n(t) x^{n+1} + \sum_{n=0}^{\infty} P_n(t) x^{n+1}-  \sum_{n=0}^{\infty} P_n(t) x^{n}$$
as $S_1+S_2+S_3+S_4+S_5$. We would like to have $x^n$ in all the sums instead of $x^{n-1}$ or $x^{n+1}$. Therefore, 
$$S_1:=\sum_{n=0}^{\infty}n P_n(t) x^{n-1}=\sum_{n=1}^{\infty}n P_n(t) x^{n-1}=\sum_{j=0}^{+\infty} (j+1)P_{j+1}(t)x^j= \sum_{n=0}^{+\infty} (n+1)P_{n+1}(t)x^n.$$ 
There is nothing to do for $S_2$ and $S_5$. Observe that 
$$S_3+S_4=\sum_ { n=0}^\infty (n+1)P_n(t)x^{n+1}=\sum_{j=1} ^{+\infty} jP_{j-1}(t)x^j=\sum_{n=1} ^{+\infty} nP_{n-1}(t)x^n.                   $$   
A: Let's expand as you did:
$$(1-2xt+x^2)\sum_{n=0}^{+\infty} nP_n(t)x^{n-1} + (x-t)\sum_{n=0}^{N} P_n(t)x^n =0 $$
$$\Leftrightarrow \sum_{n=0}^{+\infty} nP_n(t)x^{n-1} -2 \sum_{n=0}^{+\infty} ntP_n(t)x^n+\sum_{n=0}^{+\infty} nP_n(t)x^{n+1}+\sum_{n=0}^{+\infty} P_n(t)x^{n+1}-\sum_{n=0}^{+\infty} \color{red}{t}P_n(t)x^n =0$$ 
Working with each of these sums gives us :
First sum: $$\sum_{n=0}^{+\infty} nP_n(t)x^{n-1} = \sum_{n=1}^{+\infty} nP_n(t)x^{n-1}= \sum_{n=0}^{+\infty} (n+1)P_{n+1}(t)x^n= \sum_{n=1}^{+\infty} (n+1)P_{n+1}(t)x^n + P_1(t)$$
Third sum: $$\sum_{n=0}^{+\infty} nP_n(t)x^{n+1} = \sum_{n=1}^{+\infty} nP_n(t)x^{n+1} = \sum_{n=2}^{+\infty} (n-1)P_{n-1}(t)x^n = \sum_{n=1}^{+\infty} (n-1)P_{n-1}(t)x^n $$
Fourth sum: $$\sum_{n=0}^{+\infty} P_n(t)x^{n+1} = \sum_{n=1}^{+\infty} P_{n-1}(t)x^n$$
Fifth sum: $$ \sum_{n=0}^{+\infty} \color{red}{t}P_n(t)x^n = \sum_{n=1}^{+\infty} \color{red}{t}P_n(t)x^n + tP_0(t)$$
Now, if you rearrange the 5 sums, you would get
$$ \sum_{n=0}^{+\infty} nP_n(t)x^{n-1} -2 \sum_{n=0}^{+\infty} ntP_n(t)x^n+\sum_{n=0}^{+\infty} nP_n(t)x^{n+1}+\sum_{n=0}^{+\infty} P_n(t)x^{n+1}-\sum_{n=0}^{+\infty} \color{red}{t}P_n(t)x^n =0$$
$$\Leftrightarrow \sum_{n=1}^{+\infty} (n+1)P_{n+1}(t)x^n + P_1(t) -2 \sum_{n=0}^{+\infty} ntP_n(t)x^n+\sum_{n=1}^{+\infty} (n-1)P_{n-1}(t)x^n+\sum_{n=1}^{+\infty} P_{n-1}(t)x^n-\sum_{n=1}^{+\infty} \color{red}{t}P_n(t)x^n + tP_0(t)=0$$
$$\Leftrightarrow P_1(t)-tP_0(t)+\sum_{n=1}^{+\infty}((n+1)P_{n+1}(t)-2ntP_n(t)+(n-1)P_{n-1}(t)+P_{n-1}-tP_n(t))x^n =0$$
$$ \Leftrightarrow P_1(t)-tP_0(t)+\sum_{n=1}^{+\infty}((n+1)P_{n+1}(t)-(2n+1)tP_n(t)+nP_{n-1}(t))x^n =0$$
