# Introduction to Calculus equality

I was reading through Apostol's Calculus where he has this equality, but I can't figure out how to prove this equality. I would appreciate a hint if you could help. Thank you. $$[(\sum_{k=1}^{n-1}k)+1]^3-(\sum_{k=1}^{n-1}k)^3=n^3-1^3$$ Above shown to be incorrect should be: $$\sum_{k=1}^{n-1}(k+1)^3-(\sum_{k=1}^{n-1}k)^3=n^3-1^3$$

Perhaps I am wrong in conveying the equation. Let me show you where I get this from. I am trying to obtain the equation to calculate any sum $(1^2+2^2+...+n^2)$ valid for $n\in Z , n\ge1$

We start by showing $3k^2 + 3k + 1 = (k+1)^3 - k^3.$ Taking k = 1, 2, . . . , n-1 we get the n-1 formulas:

$3\bullet1^2 + 3\bullet 1 + 1 = 2^3 - 1^3$
$3\bullet2^2 + 3\bullet 2 + 1 = 3^3 - 2^3$
$3(n-1)^2+3(n-1)+1=n^3-(n-1)^3$

Now, when we add the formulas we get:

$3[1^2+2^2+. . . + (n-1)^2]+3[1+2+. . . =(n-1)]+(n-1)=n^3-1^3.$

What I don't get is by how adding the right side of the n-1 formulas we obtain the right side of the last equality.

Perhaps I wrote the sums listed above incorrectly.

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Something is wrong here. For example, if $n=3$, then $\sum_{k=1}^{n-1} k = 1+2 = 3$, and the claim is $4^3 - 3^3 = 3^3 - 1^3$. But it doesn't hold. – hardmath Feb 25 '13 at 9:32
You almost certainly mean $$\sum_{k=1}^{n-1} \big(k+1)^3 - \sum_{k=1}^{n-1} k^3 = n^3 - 1.$$ – mrf Feb 25 '13 at 9:34

I think there was a little problem with parentheses. You may intend $$\sum_{k=1}^{n-1}(k+1)^3 -\sum_{k=1}^{n-1}k^3.$$ If so, it is just a question of expanding out the sums. The first is the sum of the cubes from $2^3$ to $n^3$, and the second is the sum of the cubes from $1^3$ to $(n-1)^3$. Equivalently, it is $$\left(2^3+3^3+4^3+\cdots+(n-1)^3+n^3\right) -\left(1^3+2^3+3^3 +\cdots+(n-1)^3\right).$$ Note the wholesale cancellation when we remove the parentheses. The only things that survive are the $n^3$ from the first part and the $-1^3$ from the second part.