How Does Summing Cubic Expansions Reveal the Formula for Sum of Squares?

[sleepwalker27]

I found a deduction to determinate de sum of the first n squares. However there is a part on it that i didn't understood.


We use the next definition: $$(k+1)^3 - k^3 = 3k^2 + 3k +1$$, then we define $$k= 1, ... , n$$ and then we sum...

$$(n+1)^3 -1 = 3\sum_{k=0}^{n}k^{2} +3\sum_{k=0}^{n}k+ n$$

The left side of the equality is the one that i didn't understood. Why $$(k+1)^3 - k^3$$ changes in that way?

 

[haruspex]

((n+1)³-n³) + (n³-(n-1)³) + ((n-1)³-(n-2)³) ...

 

[Orodruin]

The first term for k is canceled by the second term for k+1. This leaves the first term for k=n and the second for k=1.

 

[Stephen Tashi]

We use the next definition: $$(k+1)^3 - k^3 = 3k^2 + 3k +1$$

A general definition is $\triangle F(k) = F(k+1) - F(k)$
So $\triangle ( k^3) = (k+1)^3 - k^3$

A general trick is the "telescoping sum":
$\sum_{k=1}^n \triangle F(k) = (F(1+1) - F(1)) + ( F(2+1) - F(2)) + (F(3+1) - F(3)) + ...+ F(n+1) - F(n)$
$= ( F(2) - F(1)) + (F(3) - F(2)) + (F(4) - F(3)) + ... + (F(n+1) - F(n))$
$= -F(1) + (F(2) - F(2)) + (F(3) - F(3)) + ...+ (F(n) - F(n))+ F(n+1)$
$= F(n+1) - F(1)$

So $\sum_{k=1}^n \triangle k^3 = (n+1)^3 - 1^3$

 

[mathman]

I found a deduction to determinate de sum of the first n squares. However there is a part on it that i didn't understood.


We use the next definition: $$(k+1)^3 - k^3 = 3k^2 + 3k +1$$, then we define $$k= 1, ... , n$$ and then we sum...

$$(n+1)^3 -1 = 3\sum_{k=0}^{n}k^{2} +3\sum_{k=0}^{n}k+ n$$

The left side of the equality is the one that i didn't understood. Why $$(k+1)^3 - k^3$$ changes in that way?

$(k+1)^3=k^3+3k^2+3k+1$. Expand the expression.