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I have to prove that this is true using mathematical induction.

I have this:

for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$

for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.

Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$

For n = $N$ + 1:

And now what? I just couldn't solve it.

grg
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1mvdb
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4 Answers4

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Perhaps the easiest way to go about this involves noticing that $$\frac 14N^4 + \frac 12N^3 + \frac 14N^2=\frac {N^2(N+1)^2}4$$

If you add $(N+1)^3$ to each side of $P(N)$ you get$$1^3+2^3+\dots N^3+(N+1)^3=\frac {N^2(N+1)^2}4+(N+1)^3$$

The left-hand side is what you need for $P(N+1)$ and the task is to show that the right-hand side can also be put into the correct form. You can extract an obvious common factor and put everything over a common denominator and see what happens.

If you don't happen to notice the convenient form for the right-hand side, you can still work with $\frac 14N^4 + \frac 12N^3 + \frac 14N^2$ to get the result.

Mark Bennet
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  • This is a new way for me, because I didn't learn to add something to P(N) to prove P(N+1), but it is very helpful and I understand it. So thank you. – 1mvdb Sep 06 '13 at 15:42
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Here's the idea. First note that your right hand side is actually $\frac{1}{4}N^2(N+1)^2$. Let's add $(N+1)^3$ to both sides of the inductive step to see if we can get $\frac{1}{4}(N+1)^2(N+2)^2$.

Doing this, we have

$$1^3 + 2^3 + \cdots + (N+1)^3 = \frac{1}{4}N^2(N+1)^2 + (N+1)^3 = (N+1)^2\left(\frac{1}{4}N^2 + N + 1\right) = \frac{1}{4}(N+1)^2(N^2+4N+4) = \frac{1}{4}(N+1)^2(N+2)^2.$$

From this, you clearly have that the truth of $P(N)$ implies the truth of $P(N+1)$. You should be able to handle write-up from here.

Cameron Williams
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Let $P(N)$ is true, that is $1^3+\ldots+n^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2$

Then for $P(N+1)$, $1^3+\ldots+n^3+(n+1)^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2+n^3+3n^2+3n+1$

$=\frac{1}{4}n^4+n^3+\frac{3}{2}n^2+n+\frac{1}{4}+\frac{1}{2}n^3+\frac{3}{2}n^2+\frac{3}{2}n+\frac{1}{2}+\frac{1}{4}n^2+\frac{1}{2}n+\frac{1}{4}$

$=\frac{1}{4}(n+1)^4+\frac{1}{2}(n+1)^3+\frac{1}{4}(n+1)^2$

That is $P(N+1)$ is true.

stcheng
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We have \begin{equation*} 1^3 + 2^3 + 3^3 + ... + N^3 +(N+1)^3= \frac 14N^4 + \frac 12N^3 + \frac 14N^2+(N+1)^3 \end{equation*} by the induction hypothesis. Also \begin{equation*} \frac{1}{4}(N+1)^4+\frac{1}{2}(N+1)^3+\frac{1}{4}(N+1)^2=\frac 14N^4 + \frac 12N^3 + \frac 14N^2+(N+1)^3 \end{equation*}

hasnohat
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