Question 1184007
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Prove by induction that for all n >= 1
1^ 5 +2^ 5 +3^ 5 +^ ...+ n ^ 5 = [n ^ 2 * (n + 1) ^ 2 * (2n ^ 2 + 2n - 1)] / 12 .
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(a) The base of induction: n = 1.

    Then the sum is one single term  {{{1^5}}},  which is 1.

    The formula (*) at n = 1 gives 

        {{{(1/12)*(1^2*(1+1)^2*(2*1^2 + 2*1-1))}}} = {{{(1/12)*(2^2*(2+2-1))}}} = {{{(1/12)*4*3}}} = {{{(1/12)*12}}} = 1,  

    so the base of induction is established.



(b)  The step of induction.

     We assume that for some integer k >= 1 this formula is valid

         1^5 + 2^5 + 3^5 + . . . + k^5 = {{{(1/12)*k^2*(k+1)^2*(2k^2+2k-1)}}}.     (1)


     We want to prove that then the formula is valid for the next integer number k+1, too:

         1^5 + 2^5 + 3^5 + . . . + k^5 + (k+1)^5 = {{{(1/12)*(k+1)^2*(k+2)^2*(2(k+1)^2+2(k+1)-1)}}}.     (2)


     At this point, the proof of the formula (2) is started.


     In the left side of (2), we replace the sum of the first k addends by the right side expression (1).

     Thus we want to prove

         {{{(1/12)*k^2*(k+1)^2*(2k^2+2k-1)}}} + {{{(k+1)^5}}} = {{{(1/12)*(k+1)^2*(k+2)^2*(2(k+1)^2+2(k+1)-1)}}}.     (3)


     Let's transform left side of (3). We factor it, taking the common factor {{{(k+1)^2}}} out of parentheses.

     Then left side of (3) takes the form

           {{{(k+1)^2 * (1/12)*k^2*((2k^2+2k-1)+12(k+1)^3)}}} = 

         = {{{(1/12)*(k+1)^2*(2k^4 + 2k^3 - k^2 + 12k^3 + 36k^2 + 36k + 12)}}} = 

         = {{{(1/12)*(k+1)^2*(2k^4 + 2k^3 + 35k^2 + 36k + 12)}}}.    (4)

    
     Now, I used an online calculator to factor an expression in the internal parentheses,
     and the calculator produced this decomposition

         {{{2k^4 + 2k^3 + 35k^2 + 36k + 12}}} = {{{(k+2)^2*(2k^2+6k+3)}}}.     (5)

     ( the link to the calculator is  https://www.pocketmath.net , the mode is "Factor" )  )


     This factorization can be continued this way
     
         {{{2k^4 + 2k^3 + 35k^2 + 36k + 12}}} = {{{(k+2)^2*(2k^2+6k+3)}}} = {{{(k+2)^2*(2(k+1)^2+2(k+1)-1)}}}.    (6)


     Now, combining all pieces (4), (5) and (6) in one whole block, we have

         {{{(1/12)*k^2*(k+1)^2*(2k^2+2k-1)}}} + {{{(k+1)^5}}} = {{{(1/12)*(k+1)^2*(k+2)^2*(2(k+1)^2+2(k+1)-1)}}}.   (7)


     It is the same as (identical to) formula (3).  Thus formula (3) is proven.



(3)  Due to the principle of the mathematical induction, it means that formula 

         {{{1^5 + 2^5 + 3^5 + ellipsis + n^5}}} = {{{(1/12)*n^2*(n+1)^2*(2n^2+2n-1)}}}.     

     is proved for all integer n >= 1.
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Solved.