Question

Sa se calculeze 1x2x3+2x3x4+...+999x1000x1001

Answer 1

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:
$S= \frac{999*1000*1001*1002}{4} =250499749500$