Swadesh

The terms in this series are $(k(k-1)+1)((k-1)!)$ . Note $1(0!)=1(1!)$ and $1(0!)+3(1!)=2(2!)$ .

Suppose the sum of the first $m$ terms is $m(m!)$ . Then the sum of the first $m+1$ terms is

$m(m!)+(m(m+1)+1)(m!)=(m^2+2m+1)(m!)$

$=(m+1)^2(m!)=(m+1)((m+1)!)$ .

Then inductively, the sum of the first $m$ terms is $m(m!)$ for any natural numbers $m$ .

In particular, the sum of the first $\boxed{2015}$ terms is $2015(2015!)$ .

see basically we can resolve the coefficients as - n^2 - n + 1 so rth term = (r-1)! * ( r^2 - r + 1 )

so open the brackets - t rth term r * r ! - r! + (r-1)! so, hence if u add terms t1 , t2 , t3 ... consecutive terms are cancelled ... left terms = 1 1! + 2 2! +3 3! ... n n! + 0! - n! but 1 * 1 ! + 2 * 2 ! ... n * n! = (n+1)! +1 ... so final results comes out to be ... n(n!) ... so answer is 2015