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what is the approach in this example:

enter image description here

I checked four inputs from zero to 3(included) and got the same output for all of them: 0,1,9,36. I guessed the law that stands behind it, which is:

enter image description here

So i proved the Base Case for n = 1, and assumed Hypothesis that n = k is true.

Solving for the Induction Step i got:

$k(k-1)(k+1)^3$

Mabadai
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    Good, for the general case you can check this https://math.stackexchange.com/a/4117390/399263. In fact the induction works a bit better with $\sum i(i+1)$ and $\sum i(i+1)(i+2)$ rather than $\sum i^2$ and $\sum i^3$ but you can convert from one another. – zwim Oct 28 '21 at 20:08

1 Answers1

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We're to show the equality for $n \geq 0$, so $n=0$ is a base case. It turns out you don't need any more base cases.

We check that the equation holds for $n = 0$. $$ \sum_{i=0}^0 i^3 = 0^3 = 0 = (0)^2 = \left( \sum_{i=0}^0 i\right)^2 \text{.} $$

Then, assuming $ \sum_{i=1}^n i^3 = \left( \sum_{i=1}^n i\right)^2 $, \begin{align*} \sum_{i=1}^{n+1} i^3 &= \left( \sum_{i=1}^n i^3 \right) + (n+1)^3 \\ &= \left( \sum_{i=1}^n i\right)^2 + (n+1)^3 \\ &= \left( \frac{n(n+1)}{2} \right)^2 + (n+1)^3 \\ &= \text{ your work goes here } \\ &= \left( \frac{(n+1)(n+2)}{2} \right)^2 \\ &= \left( \sum_{i=1}^{n+1} i\right)^2 \text{,} \end{align*} completing the induction.

Eric Towers
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