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I am interested in evaluating the following sum:

$$\chi(d,n)= d \sum_{j=0}^n (-d)^j \binom{n+2}{j+2}$$

Actually, it is not hard to evaluate this sum with a computer program, which spits out the result

$$\chi(d,n)=\frac{1}{d}((1-d)^{n+2}-1)+n+2$$

and therefore anyone interested may try to prove the above formula by induction (I tried, of course, but failed to make any real progress after applying some elementary identities relating binomial coefficients). I am hoping, however, that there is a simpler and more conceptual proof.

Some background: This sum appears as the Euler characteristic of a degree $d$ hypersurface $X_d$ in $\Bbb C\mathrm P^{n+1}$: The top Chern class is the coefficient of the $n$-th degree term in

$$c(X_d)=\frac{(1+x)^{n+2}}{(1+d x)}=(1+x)^{n+2}(1-x+x^2-x^3+\dots) $$

which clearly equals the sum I wrote down. Another factor $d$ is introduced by evaluating on the fundamental class. The above formula also appears in Dimca's book "Singularities and Topology of Hypersurfaces" (Exercise 3.7 (i) in my copy).

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It's more-or-less the binomial theorem.

$$\sum_{k=0}^{n+2}(-d)^k\binom{n+2}k=(1-d)^{n+2}$$ Delete two terms $$\sum_{k=2}^{n+2}(-d)^k\binom{n+2}k=(1-d)^{n+2}-1+(n+2)d$$ Divide by $(-d)^2$ $$\sum_{j=0}^{n}(-d)^j\binom{n+2}{j+2}=\frac{(1-d)^{n+2}-1+(n+2)d}{d^2}$$

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  • $\begingroup$ Hah, who would've known it was so simple! I clearly needed a pair of "fresh eyes" ;-) Thank you. $\endgroup$ – Danu Aug 10 '17 at 17:28

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