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