All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 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.

 See The Solution Submitted by Bractals No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 Solution | Comment 5 of 6 |

By the problem, we have:

y^2 + Sum (i= 1 to n) (y+i)^2
=  Sum (i= n+1 to 2n) (y+ 2i)^2, for n, y > 0

Substituting y = x-n, we have:

x^2 + Sum (i= 1 to n) (x-i)^2  = Sum (i= 1 to n) (x+i)^2
->  x^2 = Sum (i= 1 to n) (x+i)^2 - Sum (i= 1 to n) (x-i)^2
->  x^2 = 4x(1+2+….n)
-> x^2 = 2xn(n+1)
Thus, x = 0, or : x= 2n(n+1)
If x=0, then y = -n < 0, which is a contradiction.

Thus, x = 2n(n+1), giving: y = 2n(n+1) – n = n(2n+1).
Hence the proof.

For the sufficiency part, it can  be subsequently verified that :

(n(2n+1))^2 + Sum (i= 1 to n) ((n(2n+1))+i)^2
=  Sum (i= n+1 to 2n) ((n(2n+1))+ 2i)^2, whenever n is an integer > 0

Edited on September 27, 2008, 6:37 am
 Posted by K Sengupta on 2008-09-27 06:31:01

 Search: Search body:
Forums (0)