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**icemanfan** Problem 1 is just a straight induction. To prove that the induction holds, assume it is true for n and prove it is true for n+1. So for n+1 we add $\displaystyle (n+1)x^{n+1}$ to both sides. On the right, we need to make the addition compatible with a denominator of $\displaystyle (1-x)^2$, so multiply top and bottom by that factor and you have $\displaystyle \frac{(1-x)^2 \cdot (n+1)x^{n+1}}{(1-x)^2} = $

$\displaystyle \frac{(1 - 2x + x^2)(n+1)x^{n+1}}{(1-x)^2} =$

$\displaystyle \frac{(n+1)x^{n+1} - 2(n+1)x^{n+2} + (n+1)x^{n+3}}{(1-x)^2}$.

Now, when you add this factor on to $\displaystyle \frac{x - (n+1)x^{n+1} + nx^{n+2}}{(1-x)^2}$, you get a few cancellations. In the numerator we have:

$\displaystyle x - (n+1)x^{n+1} + nx^{n+2} + (n+1)x^{n+1} - 2(n+1)x^{n+2} + (n+1)x^{n+3}$ which, canceling like terms and combining, yields

$\displaystyle x - (n+2)x^{n+2} + (n+1)x^{n+3}$.

Hence, we have the result we want, and the induction is proven.