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.
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