a perfect square

[jay_shark]

Show that 1^3+2^3+3^3+...K^3 {sum from k=1 to n} is a perfect square for all n .

[almostbusto]

c'mon at least write up a little discussion about the problem if you are going to have someone do your homework.

[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

[Enrique]

[ QUOTE ]
Show that 1^3+2^3+3^3+...K^3 {sum from k=1 to n} is a perfect square for all n .

[/ QUOTE ]

The formula is classic.
It is ((k)(k+1)/2)^2.

There are several ways of proving it. One is induction (which you did).
The one I like is the following:
(k+1)^4 - k^4 = 4k^3 + 6k^2 + 4k + 1
k^4 - (k-1)^4 = 4(k-1)^3 + 6(k-1)^2 + 4(k-1) + 1
...
2^4 - 1^4 = 4*1^3 + 6*1^2 + 4*1^1 + 1

Summing it all you get:
(k+1)^4 - 1 = 4*sum_cubes+ 6*sum_squares + 4*sum_consecutive + k .
Now you can get the sum for cubes (assuming you have the sum of squares and the sum of consecutive formulas)

The nice thing about this technique is that if you forget the formula you can build it back up.

Sirio11 showed me another proof of it which is very pretty, but it would be hard to type the math here.

[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.