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so I recently had some lessons in induction and I feel pretty comfortable with it, but there is just this one exercise that I for some reason can't figure out.

I got this so far:

$1^3 + 2^3 + 3^3 + .... + n^3 = \frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}$ is what I need to proof, I already did the $n = 1$ and the induction hypothesis step but after that I'm stuck. I got the point where I used induction to replace the $n^3$ for the $\frac{n^4}{4} + \frac{n^3}{2} + \frac{n^2}{4}$ and made it all one big $\frac{p}{4}$ but I just can't get it to work.

If someone could help me out that would be great!

LucaMac
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L. Beute
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1 Answers1

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The base case $n = 0$ is clearly true.

Assume the following is true \begin{equation} 1^3 + 2^3 + 3^3 + \ldots + n^3 = \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 \end{equation} Let's prove \begin{equation} 1^3 + 2^3 + 3^3 + .... + n^3 + (n+1)^3 = \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 + (n+1)^3 \end{equation} The right hand side is

\begin{equation} \begin{split} \frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 + (n+1)^3 &= \frac{1}{4}n^4 + \frac{1}{2}n^3 +\frac{1}{4}n^2 + n^3 + 3n^2 + 3n + 1\\ \end{split} \end{equation} Now let's check if $$\frac{1}{4}n^4 + \frac{1}{2}n^3 + \frac{1}{4}n^2 + n^3 + 3n^2 + 3n + 1 = \frac{1}{4}(n+1)^4 + \frac{1}{2}(n+1)^3 + \frac{1}{4}(n+1)^2 \tag{1} $$ Expand each term on the RHS \begin{equation} \frac{1}{4}(n+1)^4 = \frac{1}{4}(n^4 + 4n^3 + 6n^2 + 4n + 1) \end{equation} and \begin{equation} \frac{1}{2}(n+1)^3 = \frac{1}{2}(n^3 + 3n^2 + 3n + 1) \end{equation} and \begin{equation} \frac{1}{4}(n+1)^2 = \frac{1}{4}(n^2 + 2n + 1) \end{equation} Add all the above three terms and you should get the LHS in equation $(1)$

Ahmad Bazzi
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