We want to find a formula for the sum of the cubes of the first n positive integers.
1^3+2^3+3^3+ ...+(n-1)^3+n^3
⇕
∑_(i=1)^n i^3
To do so, we will use the following to write an equation that contains the sum, ∑_(i=1)^n i^3.
i^4-(i-1)^4=4i^3-6i^2+4i-1
We sum both sides of the equation for the values of i from 1 to n.
∑_(i=1)^n(i^4-(i-1)^4)=∑_(i=1)^n(4i^3-6i^2+4i-1)
Simplifying the Left Hand-Side
Let's first simplify the left hand-side of the equation.
∑_(i=1)^n( i^4- (i-1)^4) = & 1^4- 0^4 [-0.6em]
& + 2^4-1^4
& + 3^4-2^4
& ... ...
& + n^4-(n-1)^4
∑_(i=1)^n(i^4-(i-1)^4) = & n^4
The left hand-side of the equation is equal to n^4.
∑_(i=1)^n(i^4-(i-1)^4)& =∑_(i=1)^n(4i^3-6i^2+4i-1)
n^4 & =∑_(i=1)^n(4i^3-6i^2+4i-1)
We have simplified the right hand-side of the equation. Let's review what we have done so far.
∑_(i=1)^n(i^4-(i-1)^4)& =∑_(i=1)^n(4i^3-6i^2+4i-1)
n^4 & =∑_(i=1)^n(4i^3-6i^2+4i-1)
n^4 & =4∑_(i=1)^n i^3-2n^3-n^2
Finally, we will isolate the summation notation, which represents the sum of the cubes of the first n positive integers.
