Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$ Prove that for all integers $n$, $n \geq 1$,
  $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$


How would I go about proving this?
Hint:  When $n=5$, $$\boldsymbol{1+{\color{red}3}+{\color{green}5}+{\color{purple}7}+{\color{blue}9}=} $$ $$\begin{array}{ccccc}
\blacksquare &amp; {\color{red}\blacksquare} &amp; {\color{green}\blacksquare} &amp; {\color{purple}\blacksquare} &amp; {\color{blue}\blacksquare}\\
{\color{red}\blacksquare} &amp; {\color{red}\blacksquare} &amp; {\color{green}\blacksquare} &amp; {\color{purple}\blacksquare} &amp; {\color{blue}\blacksquare}\\
{\color{green}\blacksquare} &amp; {\color{green}\blacksquare} &amp; {\color{green}\blacksquare} &amp; {\color{purple}\blacksquare} &amp; {\color{blue}\blacksquare}\\
{\color{purple}\blacksquare} &amp; {\color{purple}\blacksquare} &amp; {\color{purple}\blacksquare} &amp; {\color{purple}\blacksquare} &amp; {\color{blue}\blacksquare}\\
{\color{blue}\blacksquare} &amp; {\color{blue}\blacksquare} &amp; {\color{blue}\blacksquare} &amp; {\color{blue}\blacksquare} &amp; {\color{blue}\blacksquare}
\end{array}  $$

$$\boldsymbol{=5^2}.$$

Turn this into a general proof.
you can see this link in generally solution  :

http://en.wikipedia.org/wiki/Arithmetic_progression

This section is about Finite arithmetic series. For Infinite arithmetic series, see Infinite arithmetic series.

The sum of the members of a finite arithmetic progression is called an arithmetic series.

Expressing the arithmetic series in two different ways:

$    S_n=a_1+(a_1+d)+(a_1+2d)+\cdots+(a_1+(n-2)d)+(a_1+(n-1)d)\\
    S_n=(a_n-(n-1)d)+(a_n-(n-2)d)+\cdots+(a_n-2d)+(a_n-d)+a_n.$

Adding both sides of the two equations, all terms involving d cancel:

$    \ 2S_n=n(a_1+a_n).$

Dividing both sides by 2 produces a common form of the equation:

$    S_n=\frac{n}{2}( a_1 + a_n).$

An alternate form results from re-inserting the substitution:$ a_n = a_1 + (n-1)d:$

$    S_n=\frac{n}{2}(2a_1 + (n-1)d). $