LHS is  +...+      (1)

RHS is                     (2)





Base case:

n=1:   LHS is  = 1

       RHS is  = (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  +...+ +  

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 , giving:



LHS =  + 



    ...expand and simplify... 

    = 



    ... factor (I used WolframAlpha, you could also guess k+1 as likely

factor and do the division)...



    = 

  

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:

     = 

     = 

     =    Yes, it is the same (replace u with n).  Proof complete.



 

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