What is the proof of $n^2 = 1 + 3 + 5 ... (2*n - 1)$

Question

What is the proof of $n^2 = 1 + 3 + 5 + ... + (2\times n - 1)$?

While I verified that this is true for small numbers, I am looking for a mathematical proof for all Natural Numbers .

Answer 1

A different approach $$S=1+3+\cdots+2n-1$$ $$S=2n-1+2n-3+\cdots +1$$ adding both of them yields $$2S=2n+2n+\cdots+2n$$ implying $$2S=(2n)\cdot n $$ thus $$S=n^2$$

Answer 2

To supplement the answers already given, there is also a geometric proof of this result.

For example:

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$$ 1 + 3 + 5 + 7 + 9 = 5^2 $$

Answer 3

By induction, it is true for $n=1$, this is immediate. Suppose it is true for $n$, then let us calculate $(n+1)^2$ using the hypothesis of induction

$$(n+1)^2=n^2+2n+1=1+3+\cdots +(2n-1)+(2n+1)=1+3+\cdots +(2n-1)+(2(n+1)-1)$$ And the formula is true for $n+1$ we are done!

Answer 4

Using the summation formula of Arithmetic Series $$\sum_{r=1}^n(2r-1)=\frac n2(1+2n-1)$$

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Alternatively, $$2r-1=r^2-(r-1)^2$$

Now set $r=1,2,\cdots,n-1,n$ and add

See also: Telescoping series

Answer 5

If you already know the formula for the sum of all the positive integers between $\;1\;$ and $\;n\;$ you can do as follows:

$$1+3+5+\ldots+(2n-1)=1+2+3+4+\ldots+(2n-1)+2n-(2+4+\ldots+2n)=$$

$$=\frac{2n(2n+1)}2-2(1+2+\ldots+n)=n(2n+1)-n(n+1)=n^2$$

Answer 6

$1+3+5+\cdots+(2n-1)=\frac{(1+2n-1)\times n}{2}=n^2$