Find the sum of the sigma.

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Sequence: 1 + 2 + 3 + 4 + .... + n Sigma: n on top, i=1 and the formula is (n[n+1]) /2. 

Feb 23rd, 2015

You might want to clarify your question. n[n+1]/2 is the general equation to find the sum 1+2+3+4....+n for a given n. 

So to say add the numbers from 1-10, it's 1+2+3...+10...so n =10, so the sum is 10*(10+1)/2 = 55. But you have to be given an n to solve the sum.

But "find the sum of the sigma" alone does not make sense. If you have a follow-up question I can help!

Feb 23rd, 2015

Hi :) that was only a facet of the problem. i should have clarified but since that was a "free" question, I didn't want to make it too complicated. That was one of the formulas and sigma i was given to calculate a different sigma, which has 8 on top, i=1, and (i+i^2). So that sigma I stated in the problem was just for i. So, I was confused on how to approach it. Would I just plug in 8 to find the sum of that first sigma? Thanks! 

Feb 23rd, 2015

hi! hm I'm confused by what two sigmas are in your problem, but if you are trying to find the sum from i=1 to 8 , then yes, you plug in 8 into the n(n+1)/2 to get 8*9/2 = 36. If you want to work out another problem, feel free to invite me :)

Feb 23rd, 2015

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