Happy weekend question- 04 Sept 2010

Are we supposed to know things like triangular numbers on GMAT.
Is there any simlper way to approach this problem

Sn = \(1*2 + 2*3 + 3*4 + 4*5 ............99*100\)

= summation of n*(n+1) where n varies from 1 to 99

= summation of n^2 + n = \(n*(n+1)*\frac{(2n+1)}{6}\) \(+ n*\frac{(n+1)}{2}\)

= \(n*(n+1)*\frac{(n+2)}{3}\)
Now put n =99 we get sum = \(99*100*\frac{101}{3}\)

This is divisible by 101, hence remainder is 0

Answer is A.

Well. I guess little nuggets of such information might come in handy. And if there are more than one way to approach the problem then it is little comforting.

Before posting the answer, I had no clue as to what a triangular number is/was. I now can say that I know just a little.

[P.S -- not sure of whether to use "is or was" -- will have to focus on my SC skills]

I HAD no clue as to what a triangular number is/was => USE WAS
I HAVE no clue as to what a triangular number is/was => USE IS

HAVE > IS
HAD > WAS

Hi there Mr. Singh
I don't quite get this formula. How come you only plug 99. What does the 99 mean in this context. We basically have a geometric progression but I don't see where you plug the ratio in any part of this formula

Thank you
Cheers
J

Hi, jlgdr

I may be able to help.

In the series : S = 1*2 + 2*3 + 3*4 + 4*5 ............99*100 , first term have 1, second term have 2, third term have 3 so there is a pattern. That means 99 in last term indicates that there are total 99 terms.
So total no of terms (n) = 99

what we have here is Sn = 1*2 + 2*3 + 3*4 + 4*5 ............99*100
Or Sn = 1*(1+1) + 2*(2+1) + 3* (3+1)......... 99*(99+1)
Sn = (1+2+3+4...99) * {(1+1) + (2+1) + (3+1)......... (99+1) }
Sn = ∑n * ∑ (n+1)
Sn = ∑ n(n+1)
Sn = ∑ n + ∑n^2
Sn = n(n+1)/2 + n(n+1)(2n+1) /6
Sn = n(n+1)/2 [ 1 + (2n+1)/3 ]
Sn = n(n+1)/2 * [ 2(n+2)/3 ]

put n=99 ,
Sn = (1/2) *99 * 100 *2 *101 * (1/3)
as it is divisible by 101 , hence remainder 0..

that's my explanation, hope it helps.

Dude this is quite a big deal to solve on test day. Honestly, I would not attempt to solve such a complicated question..
But having said that I also think that this is a fairly good approach for solving given the complexity

Cheers!
J

can someone pls. explain how ∑n^2 = n(n+1)(2n+1) /6

Or Sn = 1*(1+1) + 2*(2+1) + 3* (3+1)......... 99*(99+1)
Sn = (1+2+3+4...99) * {(1+1) + (2+1) + (3+1)......... (99+1) } what happened here ? could not understand .
also , is it sub-600 level question on GMAT ? if yes then i am sure i am going to score sub-400 on GMAT.