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.

to more generalize this solution let f(n) and g(n) be two functions such that they have positive integer values for all integers n>0

then lets say we want to find what relationships f and g must have such that the sum of the first g(n)+1 squares starting at f(n) is equal to the sum of the next g(n) squares

in other words,

sum( [f(n)+t]^2 t=0 to g(n)) = sum( [f(n)+g(n)+t]^2, t=1 to g(n))

sum(f(n)^2+2f(n)t+t^2,t=0 to g(n))=sum([f(n)+g(n)]^2+2(f(n)+g(n))t+t^2,t=1 to g(n))

t^2 drops out of each side

f(n)^2(g(n)+1)+f(n)g(n)(g(n)+1)=(f(n)+g(n))^2g(n)+(f(n)+g(n))g(n)(g(n)+1)

f(n)(g(n)+1)(f(n)+g(n))=(f(n)+g(n))g(n)(f(n)+2g(n)+1)

now if f(n)=-g(n) both sides are zero, so we get f(n)=-g(n)

cancel out the f(n)+g(n)

f(n)g(n)+f(n)=f(n)g(n)+2g(n)^2+g(n)

f(n)=2g(n)^2+g(n)

f(n)=g(n)(2g(n)+1)

now if we set g(n)=n then we get f(n)=n(2n+1) and that is simply the problem given

Posted by Daniel on 2008-09-27 11:55:04