Question 1023504
{{{1^2 + 3^2 }}}+ . . . + {{{(2n - 1)^2 =(1/3)(4n^3 - n)}}}

Proof by induction.

For n = 1, the statement is trivially true.
Assume the statement is true for some natural number n = k,
or
{{{1^2 + 3^2 }}}+ . . . + {{{(2k - 1)^2 =(1/3)(4k^3 - k)}}}

Then by adding {{{(2k+1)^2}}} on both sides, we get

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

={{{(4k^3 - k + 3(2k+1)^2)/3 = (4k^3-k+12k^2+12k+3)/3 = ((4k^3+12k^2+12k+4) - k -1)/3 = (4(k+1)^3 - (k+1))/3}}}, and the statement is proved...