# What is the value of sum of k!*k, for k = 1 to n?

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Required to find the sum k!*k for k = 1 to k = n.

Let Sn be the sum of the n terms of the series.. We write the actual series for k = t to k= n:

Sn = 1!/*1+2!*2+3!*3+4!*4 +....+(n-1)!.(n-1)+n!*n-----(1).

We take the k th term k!*k= (1*2*3*....r)*k , k being an integer of natural number.

So r!*r = (1*2*3*4*...*k)(k+1 -1) = (k+1)! - k!....(2).

So now we rewrite each term of on the right side of (1) as in (2):

Sn = (2!-1!)+(3!-2!)+(4!-3!)+(5!-4!)+.......[(n-1)!-(n-2)!]+ [(n!-(n-1)!+[(n+1)!-n!].-----(3).

The right side of (3) could be rearranged as below:

Sn = -1!+{(-2!+2!)+(-3!+3!)+(-4!+4!)+......(-n!+n!)}+(n+1)!.

Sn = -1!+(n+1)!.

Sn = (n+1)!-1, as 1!= 1.

**Therefore the sum of k!*k for k= 1 to k = n is (n+1)!-1.**