Find value of $ frac{C_0}{2}-frac{C_1}{3}+frac{C_2}{4}+...+(-1)^nfrac{C_n}{n+2}, $ where $C_i = binom{n}{i} $

Question

I saw some posts in the net with solution to this, with lots of calculations on formula for combinations, wondering if it can be obtained with less calculations.
My try:
$ (1-x)^n = C_0-C_1x+C_2x^2 + ... $
$x(1-x)^n = C_0x-C_1x^2+C_2x^3 + ... $
Now integrating both sides and putting x =1, the right hand side becomes the desired expression, but I am stuck on how to calculate left hand side to get the value.

Harshit Raj · Accepted Answer

LHS=
$$int_{0}^{1}{x(1-x)^ndx} $$
$$=int_{0}^{1}{(1-(1-x))(1-x)^ndx} $$
$$=int_{0}^{1}{(1-x)^ndx} - int_{0}^{1}{(1-x)^{n+1}dx}$$
You can do rest, i believe.

Claude Leibovici · Answer

Another way to do it.
Consider
$$S=sum_{i=0}^n (-1)^i frac{  binom{n}{i}}{i+2}x^{i+2}$$
$$S'=sum_{i=0}^n (-1)^i binom{n}{i}x^{i+1}=xsum_{i=0}^n (-1)^i binom{n}{i}x^{i}=x(1-x)^n$$ and then, what @Harshit Raj showed in his/her answer.
When done, just make $x=1$ for the final result.

Find value of $ frac{C_0}{2}-frac{C_1}{3}+frac{C_2}{4}+...+(-1)^nfrac{C_n}{n+2}, $ where $C_i = binom{n}{i} $

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