1. ## Sum

Hi everyone, I didn't know where to ask this, so I thought I might just ask here:

The sum:

1/2 + (1/3+2/3) + (1/4+2/4+3/4) + (1/5+2/5+3/5+4/5) + ... + (1/100+...+99/100) is:

(A) 105 (B) 245 (C) 2475 (D) 3215 (E) 2635

Could someone tell me the fastest way of being able to simplify it, and I'm not allowed to use a calculator, thanks.

2. Originally Posted by DivideBy0
Hi everyone, I didn't know where to ask this, so I thought I might just ask here:

The sum:

1/2 + (1/3+2/3) + (1/4+2/4+3/4) + (1/5+2/5+3/5+4/5) + ... + (1/100+...+99/100) is:

(A) 105 (B) 245 (C) 2475 (D) 3215 (E) 2635

Could someone tell me the fastest way of being able to simplify it, and I'm not allowed to use a calculator, thanks.
Hello,

I've attached an image of my calculations.

EB

3. Thanks... I only have a wikipedian-based knowledge of sums, however, so although I understand the working, I don't understand how you come to that conclusion, it was on a Grade 9-10 competition paper after all...

4. Hello, DivideBy0!

The sum:

1/2 + (1/3 + 2/3) + (1/4 + 2/4 + 3/4) + (1/5 + 2/5 + 3/5 + 4/5) + ... + (1/100 + ... + 99/100)

is: .(A) 105 . (B) 245 . (C) 2475 . (D) 3215 . (E) 2635

We have: .S .= .1/2 + (1+2)/3 + (1+2+3)/4 + (1+2+3+4)/5 + ...

. . Each numerator is the nth triangular number: .½n(n + 1)

. . Each denominator is: n + 1

Hence, each term is: .½n(n+1)/(n+1) .= .½n

. . . - . - . . . .
99 . . . . . . . . . . . .99
Then: .S . = . ∑ ½n . = . ½∑ n . = . ½·½·99(100) . = . 2475
. . . . . . . . . .
n=1 . . . . . . . . . . .n=1

5. Ah, yes, thanks both of you, it made it so much clearer to form triangular numbers over the same denominator.

It's easiest for me to think of it like this: