14183

14183 (Gheorghe Szőllőssy)

Să se calculeze suma $S=\displaystyle \sum _{k=0}^{n}\left(k+1\right)^{2}C_{n}^{k}$.

Soluție Pentru orice număr natural $p$ considerăm $S(p,n)=\displaystyle \sum _{k=0}^{n}k^{p}C_{n}^{k}$. Pentru orice număr natural $n$ au loc egalitățile

$S(0,n)=\displaystyle \sum _{k=0}^{n}C_{n}^{k}=2^{n}$

$S(1,n)=\displaystyle \sum _{k=0}^{n}kC_{n}^{k}=n2^{n-1}$

$S(2,n)=\displaystyle \sum _{k=0}^{n}k^{2}C_{n}^{k}=n\left(n+1\right)2^{n-2}$.

Cum $S=\displaystyle \sum _{k=0}^{n}\left(k+1\right)^{2}C_{n}^{k}=\displaystyle \sum _{k=0}^{n}k^{2}C_{n}^{k}+2\cdot \displaystyle \sum _{k=0}^{n}kC_{n}^{k}+\displaystyle \sum _{k=0}^{n}C_{n}^{k}$, se obține $S=n\left(n+1\right)2^{n-2}+2\cdot n2^{n-1}+2^{n}$, deci $S=\displaystyle \sum _{k=0}^{n}\left(k+1\right)^{2}C_{n}^{k}=2^{n-2}\left(n^{2}+5n+4\right)$.