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Algebra Level 4

$\displaystyle \sum_{i=1}^{49}$ $i*(i+2)*(i!)^{2}$ If the sum can be expressed as $(A!)^{2} - (B!)^{2}$ where B is not equal to zero. Find A+B

49 52 50 51

1 solution

Khang Nguyen Thanh
Sep 7, 2015

We have:

$\quad\displaystyle \sum_{i=1}^{49} i(i+2)(i!)^2$

$=\displaystyle \sum_{i=1}^{49} \left((i+1)^2-1\right)(i!)^2$

$=\displaystyle \sum_{i=1}^{49} \left((i+1)!\right)^2-(i!)^2$

$=(50!)^2-(1!)^2$

Since $B\ne0$ , implies that $A=50, B=1$ , so $A+B=50+1=\boxed{51}$ .