# Telescoping sum of powers

$$\begin{array}{rclll} n^3-(n-1)^3 &= &3n^2 &-3n &+1\\ (n-1)^3-(n-2)^3 &= &3(n-1)^2 &-3(n-1) &+1\\ (n-2)^3-(n-3)^3 &= &3(n-2)^2 &-3(n-2) &+1\\ \vdots &=& &\vdots & \\ 3^3-2^3 &= &3(3^2) &-3(3) &+1\\ 2^3-1^3 &= &3(2^2) &-3(2) &+1\\ 1^3-0^3 &= &3(1^2) &-3(1) &+1\\ \underline{\hphantom{(n-2)^3-(n-3)^3}} & &\underline{\hphantom{3(n-2)^2}} &\underline{\hphantom{-3(n-2)}} &\underline{\hphantom{+1}}\\ n^3-0^3 &= & 3f_2(n) &-3f_1(n) &+n \end{array}$$

Can somebody explain me how these results are disposed intuitively? I didn't understand why $$(n-1)^3 -(n-2)^3$$ became equals to $$3(n-1)^2 - 3(n-1) + 1$$

How do this transformation was done?

Thanks!

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$(x-1)^3=x^3-3x^2+3x-1$; now set $x=n-1$.

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Look at $n^3 - (n-1)^3 = 3n^2 - 3n + 1$. Now substitute.

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Notice that $$(n-2)^3=((n-1)-1)^3=(n-1)^3-3(n-1)^2+3(n-1)-1$$

Anyway, related to the image you posted (which I didn't notice at first sight), the result was really obtained by substitution from the first row ($n\to n-1$), as other people also observed.

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An idea: since

$$A^3-B^3=(A-B)(A^2+AB+B^2)$$

we get then

$$(n-1)^3-(n-2)^3=\left[(n-1)-(n-2)\right]\left[(n-1)^2+(n-1)(n-2)+(n-2)^2\right]\ldots\ldots$$

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