Question

Hallar la suma total del siguiente arreglo:

20
19; 19
18; 18; 18
17; 17; 17; 17
.
.
.
.
.
1; 1; 1; 1; ......;1
Con procedimiento o formula que se use porfavor!

Answer 1

$\displaystyle \sigma = 1(20)+2(19)+3(18)+4(17)+...+n(21-n)+...+20(1)\\ \\ \sigma =\sum\limits_{n=1}^{20}n(21-n)\\ \\ \\ \sigma =\sum\limits_{n=1}^{20}21n-n^2\\ \\ \\ \sigma =\sum\limits_{n=1}^{20}21n-\sum\limits_{n=1}^{20}n^2\\ \\ \\ \sigma =21\sum\limits_{n=1}^{20}n-\sum\limits_{n=1}^{20}n^2\\ \\ \\ \sigma =21\times \left.\dfrac{n(n+1)}{2}\right|_{n=20}-\left.\dfrac{n(n+1)(2n+1)}{6}\right|_{n=20}\\ \\ \\ \sigma =21\times \dfrac{20(21)}{2}-\dfrac{20(21)(2(20)+1)}{6}\\ \\ \sigma= 4410-2870$

$\boxed{\sigma=1540}$