Stim asta:
[tex]\sum^n_{k=1}k= \frac{n(n+1)}{2}\\
\sum^n_{k=1}k^2=\frac{n(n+1)(2n+1)}{6} \\
\sum^n_{k=1}k^3= \frac{n^2(n+1)^2}{4} [/tex]
Suma noastra este:
[tex]\sum^n_{k=1}k(k+1)(k+2)=\sum^n_{k=1}(k^3+3k^2+2k)=\\\\
=\sum^n_{k=1}k^3+\sum^n_{k=1}3k^2+\sum^n_{k=1}2k=\\\\
=\sum^n_{k=1}k^3+3\sum^n_{k=1}k^2+2\sum^n_{k=1}k=\\\\
= \frac{n^2(n+1)^2}{4}+3 \frac{n(n+1)(2n+1)}{6}+2\frac{n(n+1)}{2}=\\\\
= \frac{n^2(n+1)^2+2n(n+1)(2n+1)+4n(n+1)}{4} =\\\\
= \frac{n(n+1)(n(n+1)+2(2n+1)+4)}4}=\\\\
=\frac{n(n+1)(n^2+2n+3n+6)}{4}=\frac{n(n+1)(n(n+2)+3(n+2))}{4}=\\\\
=\boxed{\frac{n(n+1)(n+2)(n+3)}4}}[/tex]
In cazul nostru n = 999, si putem aplica formula:
[tex]S= \frac{999*1000*1001*1002}{4} =250499749500[/tex]
