About the proof by induction that $\sum\limits_{i=1}^ni(i+1)=\frac13n(n+1)(n+2)$ for $n\geq 1$

Question

I'm trying to prove this by induction:

$$1*2 + 2*3 + 3*4 + \cdots + n(n+1) = (n(n+1)(n+2))/3.$$

I have done this so far:

Base Case: $n = 1$, works for both.

Induction Hypothesis: Let $n = k$, such that:

$$1*2 + 2*3 + 3*4 + \cdots + k(k+1) = (k(k+1)(k+2))/3.$$

Inductive Step: Try $n = k+1$:

$$1*2 + 2*3 + 3*4 + \cdots+ k(k+1) + (k+1)(k+2) = (k(k+1)(k+2))/3.$$

Apparently it's a fallacy to write $k+1$ in LHS and RHS, so I'm doing the left; however, I'm stuck at this step and not sure where to go from here.

Answer 1

It seems like you do not really understand why induction is a valid proof technique (it may also help to read this post on how to write a clear induction proof). The "fallacy" you seem to have in mind does not make a great deal of sense; the goal of induction proofs (assuming you're trying to prove a statement $S(n)$ is true) is, indeed, to move from the left-hand side of $S(k+1)$ to the right-hand side of $S(k+1)$. It does not matter from which end you start (after all, if $x=y$, then $y=x$), but the important thing is that you use the induction assumption somewhere along the way. See if you can follow the proof below, and feel free to comment if a step does not make sense.

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For $n\geq 1$, we want to show that $$ 1\cdot2+2\cdot3+\cdots+n(n+1)=\frac{n(n+1)(n+2)}{3}.\tag{1} $$ To make $(1)$ more manageable, let's use $\Sigma$-notation to rewrite it: $$ \sum_{i=1}^ni(i+1)=\frac{n(n+1)(n+2)}{3}. $$ Thus, for each $n\geq 1$, define the statement $$ S(n) : \sum_{i=1}^ni(i+1)=\frac{n(n+1)(n+2)}{3}. $$

Base step ($n=1$): The statement $S(1)$ says that $1\cdot2=2=\frac{1(2)(3)}{3}$, and this is true.

Inductive step ($S(k)\to S(k+1)$): For some $k\geq 1$, assume that $$ S(k) : \color{blue}{\sum_{i=1}^ki(i+1)=\frac{k(k+1)(k+2)}{3}} $$ holds. To be shown is that $$ S(k+1) : \color{green}{\sum_{i=1}^{k+1}i(i+1)=\frac{(k+1)(k+2)(k+3)}{3}} $$ follows. Starting with the left-hand side of $S(k+1)$, \begin{align} \color{green}{\sum_{i=1}^{k+1}i(i+1)} &= \color{blue}{\sum_{i=1}^ki(i+1)}+(k+1)(k+2)\tag{by defn. of $\Sigma$}\\[1em] &= \color{blue}{\frac{k(k+1)(k+2)}{3}}+(k+1)(k+2)\tag{by $S(k)$, ind. hyp.}\\[1em] &= \frac{\color{red}{k}(k+1)(k+2)}{3}+\frac{\color{red}{3}(k+1)(k+2)}{3}\tag{common denom.}\\[1em] &= \frac{\color{red}{(k+3)}(k+1)(k+2)}{3}\tag{combine terms}\\[1em] &= \color{green}{\frac{(k+1)(k+2)(k+3)}{3}},\tag{rearrange} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.

Thus, by induction, the claim $S(n)$ holds for all $n\geq 1$.

Answer 2

Assume that $$\text{"$P(n)$ is true" is implied by}\sum_{i=1}^n i(i+1)=\frac {n(n+1)(n+2)}3$$ First we check for the base step: $1*(1+1)=2=\frac {1*(1+1)*(1+2)}3$

Hence $P(1)$ is true. Now assume that $P(k)$ is true for some $k \in N$. So $$\sum_{i=1}^ki(i+1)=\frac {k(k+1)(k+2)}3$$ Now we check for the truth of $P(k+1)$. $$\begin{align}\sum_{i=1}^{k+1}i(i+1)&=\left(\sum_{i=1}^ki(i+1)\right)+(k+1)(k+2)\\&=\frac{k(k+1)(k+2)}3+(k+1)(k+2)\\&=(k+1)(k+2)(\frac k3+1)\\&=\frac {(k+1)(k+2)(k+3)}3\end{align}$$ This is the RHS for $P(k+1)$. Hence $P(k)$ is true $\implies$ $P(k+1)$ is true. Therefore by the principle of mathematical induction $P(n)$ is true.

Well, if you are really determined to solve the question through induction, this is the way. But there is a much easier way to prove the equality. $$\begin{align}\sum_{i=1}^ni(i+1)&=\sum_{i=1}^ni^2+i=\frac{n(n+1)(2n+1)}6+\frac{n(n+1)}2\\&=\frac16n(n+1)(2n+1+3)\\&=\frac 26n(n+1)(n+2)\\&=\frac {n(n+1)(n+2)}3\end{align}$$

Answer 3

The fallacy your professor is referring to is that you wrote down a statement that is false (even in its context).

The induction step is always to prove something of the form:

For any natural number $n$, if $P(n)$ is true, then $P(n+1)$ is true.

where $P$ is the property that you want to prove about every natural number.

You start by proving that $0$ satisfies $P$. That's fine.

But in the induction step, you are given some (unknown) natural number $n$ and you are given the fact that $n$ satisfies $P$. So while proving the induction step all you can use is $P(n)$. You must not claim $P(n+1)$ until you have proven it. It would have been okay if you had said:

Given $n \in \mathbb{N}$ such that $\sum\limits_{k=1}^n k(k+1) = \frac{n(n+1)(n+2)}{3}$, we want to prove $\sum\limits_{k=1}^{n+1} k(k+1) = \frac{(n+1)(n+2)(n+3)}{3}$, which would be equivalent to $\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$, which is true because...

This is how we can break down the task of proving the induction step, and is called a top-down approach, but you must be careful to distinguish between what you are given or have proven and what you have not proven.

However, your professor is wrong if he/she said you should only work from one side, since in more complicated problems it is usually the easiest to work from both sides and meet in the middle. This is a simple but useful technique especially if combined with the concept of 'reducing' both sides in some manner to similar form. He/she might be purposely over-simplifying the issue, though, since easy induction problems all involve proving an equality/inequality that can be done from one side to the other. Hard problems can be impossible to humanly solve without manipulating multiple pieces at the same time.