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Math proof challenge...
One night while bored, I "discovered" the following relation (turns out it's well known; not original to me):
The sum of the cubes of the first N integers is equal to the square of the sum of the first N integers. For example, take N=3: {Sum_i=1 to 3} i^3 = [{Sum_i=1 to 3}]^2 1 + 8 + 27 = [1+2+3]^2 36 = 36 Can anybody prove the general relation? |
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288?
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SO you are saying that any number is = to itself ?
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sigma[ 1^3+2^3+....+n^3 ] = (sigma [1+...+n] ) ^ 2?
i think iteration would prob be a good approach. 1^3=1^2 1^3 + 2^3 = (1+2) ^ 2 for n+1 iteration, you're adding [n+1]^3 to left. right side would be (1+2+....+n+[n+1])^2 for the equation to work. so in order for it to be true, (1+2+....+n+[n+1])^2 must equal to (1+2+...+n)^2 + [n+1]^3 going to try to factor out the [n+1] out of the squared expression (1+2+....+n+[n+1])^2 = (1+2+...+n+[n+1])*(1+2+...+n+[n+1]) = {using (a+b)*(c+d) = ac+bc+ad+bd rule} (1+2+...+n)*(1+2+...+n) + [n+1](1+2+...+n) + (1+2+...+n)*[n+1] + [n+1]*[n+1] = (1+2+...+n)^2 + 2*[n+1]*(1+2+...+n) + [n+1]^2 take step back. the first term is pulled out. (sigma(n))^2. so that means the rest must equal [n+1]^3 2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^3 {expand right} 2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^2*[n+1] {expand right} 2*[n+1]*(1+2+...+n) + [n+1]^2 ?=? [n+1]^2*n + [n+1]^2 {subtract [n+1]^2 term both sides} 2*[n+1]*(1+2+...+n) ?=? [n+1]^2*n {expand right} 2*[n+1]*(1+2+...+n) ?=? [n+1]*[n+1]*n {elimninate [n+1] factor both sides} 2*(1+2+...+n) ?=? [n+1]*n ....and i'm stuck there. who wants to finish hehe |
Fuch math....
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My cat's breath smells like cat food.
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continuing where i left off 2*(1+2+...+n) ?=? [n+1]*n
1+2+...+n = n*(n+1)/2 by Carl Gauss 2*n*(n+1)/2 ?=? [n+1]*n {cancel out 2} n*(n+1) ?=? [n+1]*n YES! finished it hehe |
I never understood that stuff in the general case.
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yea saw the typo and fixed.
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