Question

Calcular
A=1*2+2*3+3*4+...+99*100

Answer 1

Respuesta:

333300

Explicación paso a paso:

*se sabe:

$\displaystyle\sum_{i=1}^n(a_i+b_i)=\displaystyle\sum_{i=1}^na_i+\displaystyle\sum_{i=1}^nb_i$

$\displaystyle\sum_{i=1}^ni^2=\frac{n(n+1)(2n+1)}{6}$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}{2}$

*hallando la forma general de A:

$a_i=i(i+1)$

$a_i=i^{2}+i$

*desarrollando:

$\displaystyle\sum_{i=1}^n(i^{2}+i)=\displaystyle\sum_{i=1}^ni^{2}+\displaystyle\sum_{i=1}^ni$

$\displaystyle\sum_{i=1}^n(i^{2}+i)=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$

$\displaystyle\sum_{i=1}^n(i^{2}+i)=\frac{99(99+1)(198+1)}{6}+\frac{99(99+1)}{2}$

$\displaystyle\sum_{i=1}^n(i^{2}+i)=\frac{99(100)(199)}{6}+\frac{99(100)}{2}$

$\displaystyle\sum_{i=1}^n(i^{2}+i)=333300$