Prove $ \frac{0}{N} + \frac{1}{N} + \ldots + \frac{q-1}{N} =\frac{q(q-1)}{2N}$

Question

Why this equation holds? Is that by definition from a well known serie?

$$ \frac{0}{N} + \frac{1}{N} + \ldots + \frac{q-1}{N} =\frac{q(q-1)}{2N}$$

Looks like an arithmetic sequence to me. Looks like there's an obvious simplification too: factor out the common term. — , Feb 26 '13 at 12:15

score 2 · Accepted Answer · edited Apr 13 '17 at 12:21

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First we have $$ \frac{1}{N}+\frac{2}{N}+\cdots+\frac{q-1}{N}=\frac{1}{N}\left(1+2+\cdots +(q-1)\right), $$ and then we use that (see this) $$ 1+2+\cdots + (q-1)=\frac{(q-1)q}{2}. $$

edited Apr 13 '17 at 12:21

Community

1

answered Feb 26 '13 at 10:40

Stefan Hansen

25,582
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score 2 · Answer 2 · answered Feb 26 '13 at 10:41

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Your equation is a result of the following: $1+2+3...+n=\frac12n(n+1)$

We can prove this by considering, $k^2-(k-1)^2=2k-1$ and taking the sum of both sides.

answered Feb 26 '13 at 10:41

Ishan Banerjee

6,459

Prove $ \frac{0}{N} + \frac{1}{N} + \ldots + \frac{q-1}{N} =\frac{q(q-1)}{2N}$

2 Answers2