# calculate sum k!*k for k=1 to k=n

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### 2 Answers

To find sum k!*k from k = 1 to k = n.

Solution:

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

We take the rth term ar = r!*r = r1*(r+1 -r) = (r+1)! - r!.

We rewrite each term on the right side in (1) by splitting as above:

Sn = (2!-1!)+(3!-2!)+(4!-31)+(5!-4!)+....+[(n!-(n-1)!]+[(n+1)!-n1]

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

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

**Therefore the sum k!*k = (n+1)! - 1.**

Sum k!*k = 1!*1 + 2!*2 + ... + n!*n

We can add and subtract 1, so that:

Sum k!*(k + 1 - 1) = Sum k!*(k+1) - Sum k!

But k!*(k+1) = (k+1)!

We'll re-write the sum:

Sum k!*(k+1) - Sum k! = Sum (k+1)! - Sum k!

Sum (k+1)! = 2! + 3! + ... + n! + (n+1)! (1)

Sum k! = 1! + 2! + 3! + ... + n! (2)

We'll subtract (2) from (1):

Sum (k+1)! - Sum k!=2! + 3! + ..+ n! + (n+1)! - 1! - 2! - 3! - .. - n!

We'll eliminate like terms:

Sum (k+1)! - Sum k! = (n+1)! - 1

**Sum k!*k = (n+1)! - 1**