Tr = r(r+1)(r+2)(r+3)
Consider a sequence Vr such that,
Vr = r(r+1)(r+2)(r+3)(r+4)
Therefore, Vr-1 = (r-1)r(r+1)(r+2)(r+3)
Vr - Vr-1 = r(r+1)(r+2)(r+3)(-r+1 r+4)
= r(r+1)(r+2)(r+3)(5)
I.e. Vr - Vr-1 = 5Tr
So V1-V0 = 5T1
V2-V1 = 5T2
V3-V2 = 5T3 and so on till, VN - VN-1 = 5TN
Adding all these equations gives, T1+T2+.....+TN = (VN-V0)/5
Thus Sum = {N(N+1)(N+2)(N+3)(N+4)}0.2