Question

q 122. find the sum of the series given below 1(1!)+2(2!)+3(3!)+......... +2012(2012!)

Answer 1

We have to determine the sum of the series given below:

1(1!)+2(2!)+3(3!)+......... +2012(2012!)

We can express this series as $\sum _{n=1}^{n=2012}n(n!)$

= $\sum _{n=1}^{n=2012} (n+1-1)(n!)$

= $\sum _{n=1}^{n=2012} (n+1)(n!)-n!$

=$\sum _{n=1}^{n=2012} (n+1)!-n!$

$=[(1+1)!-1!]+[(2+1)!-2!]+[(3+1)!-3!]+........+[(2012+1)!-2012!]$

$=[2!-1!]+[3!-2!]+[4!-3!]+........+[2013!-2012!]$

By cancelling the terms, we get

= 2013! - 1!

= 2013! - 1

Therefore, the sum of the series is  2013! - 1.