1/1*2*3 + 1/2*3*4 + 1/3*4*5 + ......1/100*101*102* = ...... ?

Answer 1

The general term is 1/(n-1)n(n+1).

After playing with this a little, it looks like:
1/(n-1)n(n+1) =
(1/2) [1/(n-1)n - 1/n(n+1)]

So the sum is:
(1/2)[(1/1*2 - 1/2*3) + (1/2*3 - 1/3*4) + ...
+ (1/100*101 - 1/101*102)] =

(1/2)[1/1*2 - 1/101*102] =
1/4 - 1/20604

Answer 2

The sum of this series is:

S(n) = (1/4)( n(n+3) / (n+1)(n+2) )

so that for n = 100, we have

S(100) = 2575 / 10302

For the 1st three terms, or n = 3, we have

S(3) = 9 / 40 = 1 / 1*2*3 + 1 / 2*3*4 + 1 / 3*4*5

Ben's analysis is correct, and came up with the same answer and formula.

Addendum: Here's the full derivation of the formula S(n), for those that didn't follow Ben's derivation:

The first term is 1/1(1+1)(1+2) or 1/2( 1/1(1+1) - 1/(1+1)(1+2) )
The next term is 1/2(2+1)(2+2) or 1/2( 1/2(2+1) - 1/(2+1)(2+2) )
etc
The nth term is 1/n(n+1)(n+3) or 1/2( 1/n(n+1) - 1/(n+1)(n+2) )

Now, carefully notice that the right half of each term is neatly cancelled out by the LEFT half of the NEXT term. After all the cancellation, we're left with:

(1/2)( 1/(1(1+1) - 1/(n+1)(n+2) )

A few moments of simplifying this expression will get you

S(n) = (1/4)( n(n+3) / (n+1)(n+2) )

Answer 3

8O! Ben! you really know math do you? I would have answered this question since i'm already learning variables, algebra and all that stuff. I'm in 5th grade. I know what 0+a=a!!! a=any number!!!! BWAHAHA!