Sum of the series ∑r=1n r(r+1)! is
1−1n!
1−1(n+1)!
2−1(n+1)!
none of these
Let tr denote the rth term of the series, then
tr=r(r+1)!=r+1−1(r+1)!=1r!−1(r+1)!⇒∑r=1n tr=∑r=1n 1r!−1(r+1)!=1−1(n+1)!