Question

va rog sa ma ajutati ! va multumesc frumos!!
Aratati ca:
1+(1+2)+(1+2+3)+...+(1+2+3+...+n)=n(n+1)(n+2)\6

Answer 1

$\displaystyle a)\quad 1+(1+2)+(1+2+3)+...+(1+2+3+...+n)=\\ \\= \sum\limits_{k=1}^{n}\left(\sum\limits_{i=1}^{k}i\right) = \sum\limits_{k=1}^{n}\dfrac{k(k+1)}{2} =\sum\limits_{k=1}^n\dfrac{k^2+k}{2}=\\ \\ =\sum\limits_{k=1}^{n}\dfrac{k^2}{2}+\sum\limits_{k=1}^{n}\dfrac{k}{2} = \dfrac{n(n+1)(2n+1)}{2\cdot 6}+\dfrac{n(n+1)}{2\cdot 2}=\\ \\= \dfrac{n(n+1)}{2}\cdot \left(\dfrac{(2n+1)}{6}+\dfrac{1}{2}\right)=$

$=\dfrac{n(n+1)}{2}\cdot \left(\dfrac{(2n+1)+3}{6}\right)=\\ \\ =\dfrac{n(n+1)(2n+4)}{2\cdot 6} =\\ \\ = \dfrac{n(n+1)(n+2)}{6}$

$b)\displaystyle \quad 1+(1+3)+...+[1+3+5+...+(2n-1)]=\\ \\ =\sum\limits_{k=1}^n\left(\sum\limits_{i=1}^{k}(2k-1)\right) = \sum\limits_{k=1}^n\left(\dfrac{2k(k+1)}{2}-k\right) = \\ \\ =\sum\limits_{k=1}^n\left(k(k+1)-k\right)\right) = \sum\limits_{k=1}^n\left(k(k+1-1)\right) = \\ \\ = \sum\limits_{k=1}^nk^2 = \dfrac{n(n+1)(2n+1)}{6}$