perplexus.info :: Numbers : Sum of Squares

Sum of Squares (Posted on 2008-09-26)

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.

Submitted by Bractals

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Solution: (Hide)

We need to show that
SUM( i=2n²+n to 2n²+2n , i² ) = SUM( i=2n²+2n+1 to 2n²+3n , i² )
Since
SUM( i=1 to n , i² ) = n(n+1)(2n+1)/6
this reduces to showing that
SUM( i=1 to 2n²+2n , i² ) - SUM( i=1 to 2n²+n-1 , i² ) = SUM( i=1 to 2n²+3n , i² ) - SUM( i=1 to 2n²+2n , i² ) or (2n² + 2n)(2n² + 2n + 1)(4n² + 4n + 1) - (2n² + n - 1)(2n² + n)(4n² + 2n - 1) = (2n² + 3n)(2n² + 3n + 1)(4n² + 6n + 1) - (2n² + 2n)(2n² + 2n + 1)(4n² + 4n + 1)
Both sides of the above equation reduce to
n( 24n⁴ + 60n³ + 50n² + 15n + 1 )
Therefore, the proposition holds for all integers greater than zero.

Checkout Daniel's post.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
more generalized solution	Daniel	2008-09-27 11:55:04
Solution	K Sengupta	2008-09-27 06:31:01
re(3): solution (spoiler) -- print SIGMA	Daniel	2008-09-26 19:27:47
re(2): solution (spoiler) -- print SIGMA	ed bottemiller	2008-09-26 19:17:37
re: solution (spoiler)	Daniel	2008-09-26 12:39:31
solution (spoiler)	Daniel	2008-09-26 12:35:54

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