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
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© Madhur Kumar Tanwani
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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^2
n=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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