The sum of n+1 consecutive squares, beginning with the square of
n(2n+1), is equal to the sum of the squares of the next n consecutive integers.

   n=1      32 + 42 = 52
   n=2      102 + 112 + 122 = 132 + 142
      .
      .
      .

Prove that the proposition holds for all integers greater than zero.

Ówell the left hand sum is simply

Ó(n(2n+1)+t)² with t=0 to n

this simplifies to

Ó(n^2(2n+1)^2+2n(2n+1)t+t^2) with t=0 to n

n^2(2n+1)^2(n+1)+n^2(2n+1)(n+1)+n(n+1)(2n+1)/6

[6n^2(2n+1)^2(n+1)+6n^2(2n+1)(n+1)+n(n+1)(2n+1)]/6

n(n+1)(2n+1)(6n(2n+1)+6n+1)/6

n(n+1)(2n+1)(12n^2+12n+1)/6     (1)

now the right hand side is simply

Ó(n(2n+1)+n+t)^2 with t=1 to n

Ó(2n(n+1)+t)^2 with t=1 to n

Ó(4n^2(n+1)^2+4n(n+1)t+t^2) with t=1 to n

4n^3(n+1)^2+2n^2(n+1)^2+n(n+1)(2n+1)/6

[24n^3(n+1)^2+12n^2(n+1)^2+n(n+1)(2n+1)]/6

n(n+1)[24n^2(n+1)+12n(n+1)+2n+1]/6

n(n+1)[24n^3+24n^2+12n^2+12n+2n+1]/6

n(n+1)[24n^3+36n^2+14n+1]/6

n(n+1)(2n+1)(12n^2+12n+1)/6

and this is equal to (1) up above, thus this holds for all n>0

*for reference I've edited a typo I made in the simplification of the first summation, I forgot to add togeather the two 6n's to get 12n^2+12n+1 instead of 12n^2+6n+1*

Edited on September 26, 2008, 3:50 pm

Posted by Daniel on 2008-09-26 12:35:54