discrete mathematics - Induction on a sum

The left hand side has terms involving
$\binom{n}{m}= \dfrac{n!}{(n-m)!m!}$

$$1+\dfrac{1}{2}\binom{n}{1} +\frac{1}{3}\binom{n}{2}+...........+\frac{1}{n+1}\binom{n}{n} = \dfrac{2^{n+1}-1}{n+1}$$

I've done induction and proved $P(0)$ holds and also $P(1),P(2)$ holds (Just in case)

Now I've found $P(K+1)$ so the sum on the left is going to equal $\dfrac{2^{n+1}-1}{n+1}$ + $\dfrac{1}{n+2}$

and the right hand side is going to equal $\dfrac{2^{n+2}-1}{n+2}$

My problem is coming with the algebra though. I can't get those sides to equal. I can't get rid of the $n+1$ in the denominator.

Anyone have any ideas for me? They'd be much appreciated!

P.S. Someone gave me a hint and said that you can use the binomial theorem to solve this.

Answer

Recall that
$$(1+x)^n = \sum_{k=0}^n \dbinom{n}k x^k$$
Integrating this from $0$ to $1$, gives us
$$\left. \dfrac{(1+x)^{n+1}}{n+1} \right \vert_{0}^1 = \sum_{k=0}^n \dbinom{n}k \left. \dfrac{x^{k+1}}{k+1} \right \vert_0^1$$
Hence, we get what you want.