t_k  =  k(k!),  where  k = 1, 2, 3, ..... n

Sum to n terms. pls give explanations for your solution.

let s be sum to n terms,

clearly,

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

say you see there is not recurrence what so ever to solve this problem

so definitely you must think of a method which could allow you to "telescope" the terms

you observe that k(k!) is just 1(k!) less that (k+1)!

so you add and subtract k!  for all k=1,2,......n

clearly s=-1!+2!-2!.........+(n+1)!

telescoping them,we get (n+1)!-1

S_n = 1.1! + 2.2! + .............................+n.n!

S_n = (2-1)1! + (3-1)2! + ........................+ (n+1-1)n!

S_n = 2! -1! + 3! - 2! +................................+(n+1)! - n!

S_n= (n+1)! - 1