Find a formula for $\sum\limits_{r=1}^{n} (r^2+1)(r!)$

Question

The sum $$\sum\limits_{r=1}^{n} (r^2+1)(r!)$$ is equal to:

$(n+1)!$
$(n+2)!-1$
$n\cdot(n+1)!$
$n\cdot(n+2)!$

My work. I tried to solve this problem by converting $(r^2+1)$ in squares then applying the property but i was unable to get the solution, please help?

Try to use $r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2$. The sum telecopes. — Winther, Nov 28 '18 at 09:21
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. — José Carlos Santos, Nov 28 '18 at 09:23
See https://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 — lab bhattacharjee, Nov 28 '18 at 09:25
@RobertZ, I think people should get a fair idea how to address the current problem from that like https://math.stackexchange.com/questions/2638073/how-to-solve-this-summation-without-taylor/2638120#2638120 — lab bhattacharjee, Nov 28 '18 at 10:04

Robert Z · Answer 1 · 2018-11-28T10:00:39.043

4

Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment: $$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$ then $$\begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\\ &=((r+2)! - (r+1)!) - 2((r+1)!-r!). \end{align}$$ Now split the sum in two parts and try to simplify. What do you obtain?

edited Nov 28 '18 at 10:00

answered Nov 28 '18 at 09:31

Robert Z

145,942

does this form an telescoping series? – Himanshu Shekhar Nov 28 '18 at 09:49
$$f(r+2)-3f(r+1)+2f(r)=f(r+2)-f(r+1)-2{f(r+1)-f(r)}$$ – lab bhattacharjee Nov 28 '18 at 09:51
Yes, it is a telescopic sum. – Robert Z Nov 28 '18 at 10:01

Find a formula for $\sum\limits_{r=1}^{n} (r^2+1)(r!)$

1 Answers1