Question 1183221
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LHS is  {{{1^2+3^2}}}+...+{{{ (2n-1)^2 }}}      (1)
RHS is  {{{n(4n^2-1)/3}}}                   (2)


Base case:
n=1:   LHS is {{{ (2(1)-1)^2 }}} = 1
       RHS is {{{ 1*(4*1^2-1)/3 }}} = (4-1)/3 }}} = 1 

Base case holds.

Hypothesis:
       Assume LHS = RHS for  n=k             (*)


Step case: 
        Let n=k+1  (recall the index k counts by 1 and the 2k-1 in the LHS & RHS is what makes sure you have odd numbers only)

What you need to do now, is show LHS=RHS for n=k+1, then the proof is complete.

LHS is  {{{ green( 1^2+3^2 ) }}}+...+{{{ green((2k-1)^2) }}} + {{{ (2(k+1)-1)^2 }}} 
Where I have separated the (k+1)th term.  The terms in green are the n=k case, which by the hypothesis (*), can be replaced by {{{k(4k^2-1)/3}}}, giving:

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

    ...expand and simplify... 
    = {{{ (4k^3+12k^2+11k+3)/3 }}}

    ... factor (I used WolframAlpha, you could also guess k+1 as likely
factor and do the division)...

    = {{{ (2k+1)(2k+3)(k+1) / 3 }}}
  
Is this last expression the same as (2)?
  Let u=k+1,  --> k=u-1
  Then the last expression above, in terms of u, is:
     = {{{ (2(u-1)+1)(2(u-1)+3)u)/3 }}}
     = {{{ ((2u-1)(2u+1)u)/3 }}}
     = {{{  ((4u^2-1)u)/3 }}}   Yes, it is the same (replace u with n).  Proof complete.