Question 252353: sum of the first 100 terms for the series:
(1*2*3)+(2*3*4)+(3*4*5)+.....................
Found 2 solutions by kensson, palanisamy:
Answer by kensson(21)   (Show Source):
You can put this solution on YOUR website! Wow, good question!
I got 26,527,650 using Pascal's triangle and some shortcuts.
Every term is a multiple of 6 (it must have at least one multiple of 3 and an even number in the product) so I started by dividing those out. The sequence that came out was:
1, 4, 10, 20, 35, ...
which I recognised as the 4th diagonal of Pascal's triangle. The sequence of their sums is 1, 5, 15, 35, 70, ..., which is the 5th diagonal.
The sum to 100 terms is the 100th term of that sequence, which is 103C99 = 103C4 = 4,421,475. But remember, we have to multiply that by six to get 26,527,650.
Answer by palanisamy(496)  (Show Source):
You can put this solution on YOUR website! Lset S = (1*2*3)+(2*3*4)+(3*4*5)+.....................100 terms
The nth term Tn = n*(n+1)*(n+2)
= n*(n^2+2n+n+2)
= n*(n^2+3n+2)
= n^3+3n^2+2n
Sum to n terms = sum( n^3 +3n^2+2n, i=1, n )
= n^2(n+1)^2 /4 + 3 n(n+1)(2n+1)/6 + 2n(n+1)/2
Put n = 100
Sum to 100 terms = (100)^2(101)^2 /4 +3x100x(101)x(201)/6 +100(101)
= 25502500+1015050+10100
= 26527650
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