Question 1149333
<pre>
n=1:  {{{1^2 = 1 }}}   &   {{{(1/3)(1)(4(1)^2-1) = (1/3)(3) = 1 }}}

Assume {{{1^2+2^2 }}} + ... + {{{ (2n-1)^2 }}} = {{{(1/3)(n)(4n^2-1)}}} for n=k

Let n=k+1:

 {{{1^2+2^2 }}} + ... + {{{ (2k-1)^2 }}}  +  {{{ (2(k+1)-1)^2 }}}

// Apply hypothesis to all but the last term 
= {{{ (1/3)(k)(4k^2-1) }}} + {{{ (2(k+1)-1)^2 }}}

// combine
= {{{ (1/3)(4k^3+12k^2+11k+3) }}}

// Here you can attempt to factor, or, you can see if (1/3)(k+1)(4(k+1)^2-1)
// expands to the same expression...  I will do the latter approach...

{{{ (1/3)(k+1)(4(k+1)^2-1) }}}
= {{{ (1/3)(k+1)(4k^2+8k+4-1) }}}
= {{{ (1/3)(k+1)(4k^2+8k+3) }}}
= {{{ (1/3)(4k^3+8k^2+3k+4k^2+8k+3) }}}
= {{{ (1/3)(4k^3+12k^2+11k+3) }}}

They match, the hypothesis is also true for n=k+1, DONE.