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

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MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,661
St. Augustine, FL.
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.
 
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MarkFL

Pessimist Singularitarian
Staff member
Feb 24, 2012
13,661
St. Augustine, FL.
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.