Sum of odd integers

**OReilly** · May 28th 2006, 08:54 AM

How to prove that $1+3+5+...+n = \frac{(n+1)^2}{4}$ ?

**rgep** · May 28th 2006, 12:25 PM

In a word: induction.

In a thousand words:

Code:

**Soroban** · May 28th 2006, 05:32 PM

Hello, OReilly!

Here's one way . . .

How to prove that: $1+3+5+...+n = \frac{(n+1)^2}{4}$ ?

This is an arithmetic series
. . with first term $a = 1$ , common difference $d = 2$ , and $\frac{n+1}{2}$ terms

Using the Sum formula: . $S \:= \:\frac{1}{2}\cdot\frac{n+1}{2}\left[2\cdot1 + 2\cdot\left(\frac{n+1}{2} - 1\right)\right]$

. . . $= \;\frac{n+1}{4}[2 + (n + 1 - 2)] \;= \;\frac{n+1}{4}(n+1) \;= \;\frac{(n+1)^2}{4}$

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