# Vu's question at Yahoo! Answers: Inductive proof - sum of the cubes of the first n natural numbers

#### MarkFL

##### Pessimist Singularitarian
Staff member
Here is the question:

Prove that 13+23+...+n3=(n(n+1)/2)2 by using induction.?

help me
I have posted a link there to this question so the OP can view my work.

#### MarkFL

##### Pessimist Singularitarian
Staff member
Hello Vu,

We are given to prove by induction:

$$\displaystyle \sum_{k=1}^n\left(k^3\right)=\left(\frac{n(n+1)}{2}\right)^2$$

First, we must check to see if our base case $P_1$ is true:

$$\displaystyle \sum_{k=1}^1\left(k^3\right)=\left(\frac{1(1+1)}{2}\right)^2$$

$$\displaystyle 1^3=1^2$$

$$\displaystyle 1=1$$

The base case is true, so we next state the induction hypothesis $P_n$:

$$\displaystyle \sum_{k=1}^n\left(k^3\right)=\left(\frac{n(n+1)}{2}\right)^2$$

As our inductive step, we may add $(n+1)^3$ to both sides:

$$\displaystyle \sum_{k=1}^n\left(k^3\right)+(n+1)^3=\left(\frac{n(n+1)}{2}\right)^2+(n+1)^3$$

On the left, incorporate the new term within the sum and factor on the right:

$$\displaystyle \sum_{k=1}^{n+1}\left(k^3\right)=(n+1)^2\left(\left(\frac{n}{2}\right)^2+(n+1)\right)$$

$$\displaystyle \sum_{k=1}^{n+1}\left(k^3\right)=(n+1)^2\left(\frac{n^2+4n+4}{4}\right)$$

$$\displaystyle \sum_{k=1}^{n+1}\left(k^3\right)=\frac{(n+1)^2(n+2)^2}{4}$$

$$\displaystyle \sum_{k=1}^{n+1}\left(k^3\right)=\left(\frac{(n+1)(n+2)}{2}\right)^2$$

$$\displaystyle \sum_{k=1}^{n+1}\left(k^3\right)=\left(\frac{(n+1)((n+1)+1)}{2}\right)^2$$

We have derived $P_{n+1}$ from $P_n$ thereby completing the proof by induction.