The sum of n+1 consecutive squares, beginning with the square of
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.
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