Maths problem, series & series

DtotheK

Probably isn't thread worthy but anyway:

I'm stuck on this question and would be grateful for an explanation of the solution..

Consise maths 5 the blue book , page 228 chapter 7.9 question 17.

The first part is fine.
Second part

Evaluate : 1 + (1+2)+(1+2+3)+...+(1+2+3+...+20)

It's like the Sn of 20 Sn's.. I get the wrong answer and confused everytime i try it. I 'm guessing there's some reasonably quick way of doin this and not adding up the 20 Sns.

answer = 1540...
I just need to know how.!!

Thanks!!

Choco

You know, I can't work out a fancy way to do this at all. Obviously there should be a 'quick' formula-utilising way to do this, but seemingly, a bit of calculator work is the only way I can think of.

So just put this (because there are 20 1s and 19 2s etc)...

(1x20)+(2x19)+(3x18)+(4x17)+...+(19x2)+(20x1)

...in to your calculator, and after the brain-numbing task of putting it in and pressing equal, you do get the answer of 1,540 in a completely non-mathematical way.

DtotheK

AH i see !! that's a relief i was all over the shop tryin the Sn of arithmetic series and so on formulas but that puts my mind at ease!!! Cheers!

If anyone found a different disgracefully hard way of doin this drop us a dingle

JC 2K3

Every term is a sum of integers, ie:

Un = (n(n+1))/2 = (1/2)(n^2 + n)

Considering sigma n^2 = (n(n+1)(2n+1))/6

Sn = (1/2)(((n(n+1)(2n+1))/6 + (n(n+1))/2)
S20 = (1/2)(((20(21)(41))/6 + (20(21))/2)
= 1540