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We have to evaluate the following:

$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$

Any hints ?

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The result should be $2 - (n+2)! + n(n+2)!$. You can show this by induction, although I'm not sure how you can derive this result. – George V. Williams Mar 19 '13 at 0:53
up vote 8 down vote accepted


$$f_k = (k-1) (k+2)!$$


$$f_k-f_{k-1} = k^2 (k+1)!$$


$$\begin{align}\sum_{k=1}^n k^2 (k+1)! &= f_1 - f_0 + f_2 - f_1 + \ldots+f_{n}-f_{n-1}\\ &= f_n-f_0\\ &= (n-1)(n+2)! + 2\end{align}$$


How did I get my $f_k$? By assuming that such an anti-difference takes the form

$$f_k = (k+m) (k+n)!$$

This general form will produce a quadratic times some factorial upon differencing. Here

$$f_k - f_{k-1} = [(k^2-(m+n-1) k + m (n-1) + 1] (k+n-1)!$$

We want to isolate a factor of $k^2$ in the brackets. Then we have to solve

$$m+n-1=0$$ $$m (n-1) + 1=0$$

The solutions are $(m,n) = (-1,2)$ and $(1,0)$. The former solution provides the form we seek.

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It would be good to comment on how to obtain the "discrete primitive" $f_k$ in this case. – Pedro Tamaroff Mar 19 '13 at 1:44
Thanks a lot @Ron Gordon – Vishwesh Mar 19 '13 at 1:49
@PeterTamaroff: I hate to say it, but it was just a lot of messing around until I got the $f_k$ whose difference was the term being summed. I am not aware of a general method by which one finds an "anti-difference." – Ron Gordon Mar 19 '13 at 2:03
@RonGordon See Sasha's answer. – Did Mar 19 '13 at 8:00
@Did: I did, and from my perspective, it has just as much deux ex machina as mine. – Ron Gordon Mar 19 '13 at 8:02

Using $k^2 = (k+3)(k+2) - 5 (k+2) + 4$ we write $$f_k = k^2 (k+1)! = (k+3)! - 5 (k+2)! + 4 (k+1)! = G_{k+1} - G_k$$ where $G_k = (k+2)! - 4(k+1)!$, therefore $$ \sum_{k=1}^n f_k = \sum_{k=1}^n \left(G_{k+1} - G_k\right) = G_{n+1} - G_1 $$ Since $G_1 = -2$, and $G_{n+1} = (n+3)! - 4(n+2)! = (n+2)! (n-1)$ we get $$ \sum_{k=1}^n k^2 (k+1)! = (n-1) (n+2)! + 2 $$

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Using k2=(k+3)(k+2)−5(k+2)+4 we write fk=k2(k+1)!=(k+3)!−5(k+2)!+4(k+1)!=Gk+1−Gk where Gk=(k+2)!−4(k+1)!, therefore ∑k=1nfk=∑k=1n(Gk+1−Gk)=Gn+1−G1 Since G1=−2, and Gn+1=(n+3)!−4(n+2)!=(n+2)!(n−1) we get ∑k=1nk2(k+1)!=(n−1)(n+2)!+2

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Please use $\LaTeX$ when typesetting math. Use this guide:… – Dennis Gulko Mar 19 '13 at 8:17
This appears to be a duplicate of Sasha's solution, non-LaTeX'ed, but character-for-character. – Ron Gordon Mar 19 '13 at 13:28
Please do not plagiarize. This can lead to suspension. – robjohn Mar 19 '13 at 18:05

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