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For any integer $n\geq2$, prove that $\displaystyle\sum\limits_{i=1}^{n-1}i(i+1) = \frac{n(n-1)(n+1)}{3}.$

Let $P(n)$ be the formula $\displaystyle\sum\limits_{i=1}^{n-1}i(i+1) = \frac{n(n-1)(n+1)}{3}.$

$P(2)$ is true since $\displaystyle\frac{2\cdot1\cdot3}{3}=2.$

$P(k)$ gives $\displaystyle\sum\limits_{i=1}^{k-1}i(i+1) = \frac{k(k-1)(k+1)}{3}.$

We wish to show that $P(k+1)$ = $\displaystyle\sum\limits_{i=1}^{k}i(i+1) = \frac{k(k+1)(k+2)}{3}$ is true.

From the left hand side,

\begin{align}\displaystyle\sum\limits_{i=1}^{k}i(i+1) &= \displaystyle\sum\limits_{i=1}^{k-1}i(i+1) + (k+1)((k+1)+1) \\ &=\displaystyle\frac{k(k-1)(k+1)}{3}+(k+1)(k+2) \\ \end{align}

I'm pretty sure I messed up somewhere already, as after all the algebra, I can't have show it of the form $P(k+1)$. How do you these types of inductions? Where is my error? Thanks.

Martin Sleziak
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2 Answers2

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You are adding $(k+1)(k+2)$ while you should be adding $k(k+1)$. You are just messing up with the indices. There is a proof without using induction.

$$S(k)=\sum_{i=1}^{k}i(i+1)=\sum_{i=1}^{k}i^2+\sum_{i=1}^{k}i$$

and we know $\sum_{i=1}^{k}i^2=\frac{k(k+1)(2k+1)}{6}$ and $\sum_{i=1}^{k}i=\frac{k(k+1)}{2}$. (Try to prove these without using induction). So adding the two one gets

$$S(k)=\frac{k(k+1)(2k+1)+3k(k+1)}{6}=\frac{k(k+1)(2k+4)}{6}=\frac{k(k+1)(k+2)}{3}$$

marwalix
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Be careful the term of index $k$ of the sum $\sum_{i=1}^k i(i+1)$ is $k(k+1)$. Otherwise, you're in the right track.

user37238
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