a perfect square

[jay_shark]

lol @homework .

I already know the solution to this problem , I posted it here because I thought it was pretty interesting .
There is nothing wrong with posting a math problem for the sake of creating an interesting discussion .

[51cards]

[ QUOTE ]
lol @homework .

I already know the solution to this problem , I posted it here because I thought it was pretty interesting .
There is nothing wrong with posting a math problem for the sake of creating an interesting discussion .

[/ QUOTE ]

I am curious to see. Show me.

[jay_shark]

ok here is the solution :



1^3= 1^2
Take 1^3+2^3 = 3^2 this is a perfect square
Now take 1^3+2^3+3^3= 6^2 this is also a perfect square .
1^3+2^3+3^3+4^3= 10^2
One solution may be to guess what the perfect square may be in terms of n .

Notice that the summation for n=1,2,3,4 respectively is 1^2,3^2, 6^2, 10^2 . One may guess through trial and error , that the summation of the first n cubes is
C(n+1,2) . Now you may prove using induction that this does in fact hold for all n . It's true for n=1,2,3 so suppose it's true for the first k cubes . That is , 1^3+2^3+...K^3 = C(K+1,2)^2. One needs to show that if it's also true for k+1 .

1^3+2^3+...K^3+(K+1)^3 = c(K+1,2)^2 + (k+1)^3 .
Now hopefully the rhs will become equal to c(K+2,2)^2 . That is , we need to show c(K+1,2)^2 + (k+1)^3 = c(K+2,2)^2

(K+1)^2K^2/4 + (K+1)^3 = (K+1)^2[k^2/4 +(K+1)]
I factored out (K+1)^2
Now [K^2/4 + K+1] = (K+2)^2/4 and we're now finished .
Since (K+1)^2(K+2)^2/4 is exactly n(K+2,2)^2 . QED

[BruceZ]

[ QUOTE ]
ok here is the solution :



1^3= 1^2
Take 1^3+2^3 = 3^2 this is a perfect square
Now take 1^3+2^3+3^3= 6^2 this is also a perfect square .
1^3+2^3+3^3+4^3= 10^2
One solution may be to guess what the perfect square may be in terms of n .

Notice that the summation for n=1,2,3,4 respectively is 1^2,3^2, 6^2, 10^2 . One may guess through trial and error , that the summation of the first n cubes is
C(n+1,2) . Now you may prove using induction that this does in fact hold for all n . It's true for n=1,2,3 so suppose it's true for the first k cubes . That is , 1^3+2^3+...K^3 = C(K+1,2)^2. One needs to show that if it's also true for k+1 .

1^3+2^3+...K^3+(K+1)^3 = c(K+1,2)^2 + (k+1)^3 .
Now hopefully the rhs will become equal to c(K+2,2)^2 . That is , we need to show c(K+1,2)^2 + (k+1)^3 = c(K+2,2)^2

(K+1)^2K^2/4 + (K+1)^3 = (K+1)^2[k^2/4 +(K+1)]
I factored out (K+1)^2
Now [K^2/4 + K+1] = (K+2)^2/4 and we're now finished .
Since (K+1)^2(K+2)^2/4 is exactly n(K+2,2)^2 . QED

[/ QUOTE ]

That's how I did it too. Note that C(K+1,2) is also known as the sum of the first K positive integers. So we have proven that for any K, the sum of the first K consecutive cubes is equal to the square of the sum of the first K consecutive positive integers. That is:

1^3 + 2^3 + 3^3 + 4^3 + ... + K^3 = (1 + 2 + 3 + 4 + ... + K)^2.