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

Question

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 ?

Answer 1

Let

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

Then

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

and

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

ADDENDUM

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.

Answer 2

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 $$ \sum_{k=1}^n k^2 (k+1)! = (n-1) (n+2)! + 2 $$

Answer 3

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