Question

Salut,am nevoie de ajutor la ex. 1.339B

Answer 1

$\displaystyle\sum\limits_{k=1}^{2n}(-1)^kk^2=-1^2+2^2-3^2+4^2-...+(2n)^2 = \\ \\ = \displaystyle \sum\limits_{k=1}^{n}\Big[-(2k-1)^2+(2k)^2\Big]=\sum\limits_{k=1}^{n}\Big[(2k)^2-(2k-1)^2\Big] = \\ \\ = \sum\limits_{k=1}^{n}\Big[(2k-2k+1)(2k+2k-1)\Big] = \\ \\ = \sum\limits_{k=1}^{n}(4k-1)= \\ \\\\ = \dfrac{4n(n+1)}{2} - n=\dfrac{4n(n+1)-2n}{2} = 2n(n+1)-n=\\ \\ = 2n^2+n = n(2n+1)$

=> a) corect.