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

Mathematics Asked by Monalisha Bhowmik on November 11, 2020

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.

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.

Correct answer by Harshit Raj on November 11, 2020

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.

Answered by Claude Leibovici on November 11, 2020

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