How do I prove the equality of two sums?

Question

I'm trying to prove the following:

$$\sum_{r=0}^{n+1} \frac{n! (n+1)}{r!(n-r)!(n-r+1)} = \sum_{r=0}^n \frac{2n!}{r!(n-r)!}$$

I'm pretty sure that these are equal. How could I go about proving it? Thank you.

Answer 1

Let $$\displaystyle S = \sum^{n+1}_{r=0}\frac{n!\cdot (n+1)}{r!\cdot (n-r)!\cdot (n-r+1)} = \sum_{r=0}^{n+1}\frac{(n+1)!}{(n-r+1)!\cdot r!} = \sum^{n+1}_{r=0}\binom{n+1}{r}.$$

We have $$(1+x)^s = \binom{s}{0}+\binom{s}{1}x+\binom{s}{2}x^2+.........+\binom{s}{s}x^n$$ due to the binomial theorem.

Now put x=1 and s=n+1. We get $$ 2^{n+1} = \sum^{n+1}_{r=0}\binom{n+1}{r}.$$

And for x=1 and s=n, we get $$ 2^n = \sum^{n}_{r=0}\binom{n}{r}.$$

Hence, we get $$\displaystyle S=\sum^{n+1}_{r=0}\binom{n+1}{r} = 2^{n+1} = 2\cdot 2^n = 2\sum^{n}_{r=0}\binom{n}{r} = \sum^{n}_{r=0}\frac{2\cdot n!}{r!\cdot (n-r)!}.$$

Answer 2

HINT:

$$\dfrac{(n+1)\cdot n!}{r!(n-r)!\cdot(n-r+1)}=\dfrac{(n+1)!}{(n+1-r)! r!}$$

Set $n+1=m$