Find the value of: $\sum_{n=1}^{50}n(n!)$

Question

Find the value of the summation: $$\sum_{n=1}^{50}n(n!)$$

The solution in the answer key is $51!-1$. I am unable to find the given solution.

Thanks in advance!

Answer 1

Hint

• Use $n \cdot n! = (n+1 - 1) \cdot n! = (n+1)! - n!$

Answer 2

Note that with just a little experimentation you can guess the result. Calculate

$$\sum_{k=1}^nk\cdot k!$$

for $n=1,2,3$, and $4$, and you get $1,5,23$, and $119$. If necessary, you can go one step further and for $n=5$ get $719$. Those numbers should be recognizable as $2!-1,3!-1,4!-1,5!-1$, and $6!-1$. Now even if you don’t see the telescoping trick, you can guess that

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

and try to prove it by induction, something that turns out to be very easy.