Question

1(1!)+2(2!)+3(3!)....2012(2012!) what is the sum of the series? in terms of n! and constants.

Answer 1

We have to evaluate the sum of the given series

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

It can be generalized as $\sum _{n=1}^{2012} n(n!)$

Now, expressing n as (n+1)-1

Therefore, the series is $\sum _{n=1}^{2012} (n+1-1)(n!)$

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

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

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

= (2!-1!)+(3!-2!)+(4!-3!)+.....+(2013!-2012!) [/tex]

Cancelling the terms, we get

= 2013! - 1

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