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

to both sides. On the right, we need to make the addition compatible with a denominator of

, so multiply top and bottom by that factor and you have

.

Now, when you add this factor on to

, you get a few cancellations. In the numerator we have:

which, canceling like terms and combining, yields

.

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