Problem involving Integers and Proof

Question

Compute the following sums: a) 1 + 2 + 3 + ... + n; b) 1 + 3 + 5 + ... + 2n-1; c) 2 + 5 + 8 + ... + 3n-1 d) a + (a+d) + (a+2d) + ... + (a+(n-1)d)

I have found that the sum of a) = -1/12. I have no idea how to find the rest, does anyone else have any idea? Any help is much appreciated

To answer you questions see See http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF#Partial_sums — user127.0.0.1, Feb 18 '14 at 21:04

score 0 · Answer 1 · edited Apr 13 '17 at 12:20

0

These are all presumably finite sums, not infinite. These are all sums of arithmetic progressions. The first should be $\frac 12n(n+1)$. Presumably your answer of $-\frac 1{12}$ was found by Google search, I suggest you unlearn it. It was discussed here

edited Apr 13 '17 at 12:20

Community

1

answered Feb 18 '14 at 21:06

Ross Millikan

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why is it 1/2n(n+1) that doesn't seem to make sense because this seems to result in, (1,3,6,10...) – Roy Kesserwani Feb 18 '14 at 21:13
Yes, and that is the sum of $1,2,3,\dots$ terms of the progression. For example, $1+2+3=6$ – Ross Millikan Feb 18 '14 at 21:17

Problem involving Integers and Proof

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