More formally, find a general solution for :

**n**

**Σ (2k-1)**

**k=1**

Source : CLRS A.1-1

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#
Find the summation of first 'n' odd numbers

## Labels / Tag

Find a simple formula for the summation of the first 'n' odd numbers (1, 3, 5.....).

More formally, find a general solution for :** **

**n**

**
Σ (2k-1)**

**
k=1**

Source : CLRS A.1-1

More formally, find a general solution for :

Source : CLRS A.1-1

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## 3 comments:

A solution for this is available on the mathforum : http://mathforum.org/library/drmath/view/56886.html

However, I find this proof easier. I'll use "S" instead of the sigma sign. All ranges are from 1 to n.

Using linearity property :

S(2k-1) = 2S(k) - S(1)

We know that S(k) is n(n+1)/2, hence :

S(2k-1) = 2(n)(n+1)/2 - n

which means :

S(2k-1) = n^2n=1 => sum=1

n=2 => sum=4

n=3 => sum=9

If you want to find the sum of first n odd numbers, then the answer is n*n. It's that simple

@Madan

That's correct.

The way to derive that the summation is n^2, is to either assume it is n^2 and then prove it via induction or any of the other proofs stated earlier.

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