Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am stuck at this step in the inductive process and I was wondering if someone can help me out from where I am stuck.


if $n$ is a positive integer, prove that,

$$1\cdot2+2\cdot3+3\cdot4+\cdots+n\cdot(n+1) = \frac{n(n+1)(n+2)}3$$

Basic step is $1$, then assume $k$ is true and we get,

$$1\cdot2+2\cdot3+3\cdot4+\cdots+k\cdot(k+1) = \frac{k(k+1)(k+2)}3$$

After that I am stuck because the answer key is telling me that the next step is to add $(k+1)(k+2)$, but I don't know why we do that. Aren't we supposed to just plugin $k+1$ into all $k$ variables. Why are we adding $(k+1)(k+2)$ to both sides?

share|cite|improve this question
up vote 3 down vote accepted

So we want to prove that

$$1\cdot2+2\cdot3+3\cdot4+\cdots+n(n+1)=\sum_{i=1}^{n}{i(i+1)}=\frac{n(n+1)(n+2)}{3}\forall n\in\mathbb{N}$$

The base case ($n=1$):

Note that $1\cdot2=2$ and that $\frac{1\cdot2\cdot3}{3}=2$. Thus the base case holds.

Suppose that $\exists k\in\mathbb{N}$ such that $\sum_{i=1}^{k}{i(i+1)}=\frac{k(k+1)(k+2)}{3}$. Now consider the case that $n=k+1$:


By the induction hypothesis (we claimed that for $n=k$ the formula held) we have:


Therefore, by induction, the formula is true $\forall n\in\mathbb{N}$

share|cite|improve this answer
How come you add ∑i=1ki(i+1) at the k+1 part? – user3718584 Jul 3 '14 at 19:23
I split up the sum to $k+1$. I split it into the sum up to $k$ and then the $(k+1)^{\text{st}}$ term, which is $(k+1)(k+2)$. I split the sum up so I can apply the induction hypothesis, that the sum up to $k$ is equal to $\frac{k(k+1)(k+2)}{3}$. Does that make sense? – dleggas Jul 3 '14 at 19:32
Yup thanks for clarifying – user3718584 Jul 3 '14 at 19:48

Hint $\: $ It is very easy to inductively prove the following basic theorem on additive telescopy

$$\rm\ g(n)\ =\, \sum_{i\: =\: 1}^n\:\ f(i)\ \iff\ \ g(n) - g(n\!-\!1)\ =\ f(n)\ \ for\ \ n> 1,\,\ \ g(1)=f(1)$$

Your is special case $\rm\ f(n) = n(n\!+\!1),\ \ g(n) = n(n\!+\!1)(n\!+\!2)/3,\, $ which is easily checked to satisfy the above. Indeed, $\rm\,g(n)-g(n\!-\!1) = n(n\!+\!1)((n\!+\!2)-(n\!-\!1))/3 = n(n\!+\!1)\ $ and $\rm\ g(1) = f(1) $

The inductive proof of the general theorem is easier than that for your special case because the cancellation that occurs is much more obvious at this level of generality, whereas it is usually obfuscated by details of specific instances. Namely, the proof of the general theorem is just a rigorous inductive proof of the following telescopic cancellation

$$\rm \underbrace{\overbrace{g(1)}^{\large f(1)}\phantom{-g(1)}}_{\large =\ 0}\!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{-\,g(1)\!+\!\phantom{g(2)}}^{\large f(2)}\!\!\!\!\!\!\!\!\! \underbrace{g(2) -g(2)}_{\large =\ 0}\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overbrace{\phantom{-g(2)}+ g(3)}^{\large f(3)}\!\!\!\!\!\!\!\!\!\!\underbrace{\phantom{g(3)}-g(3)}_{\large =\ 0}\!+\:\cdots +\!g(n)\ =\ g(n) $$

This theorem reduces the inductive proof to simply verifying the RHS equalities, which is trivial polynomial arithmetic when $f(n),g(n)$ are polynomials in $n$. The above theorem is an example of telescopy, also known as the Fundamental Theorem of Difference Calculus, depending on context. You can find many more examples of telescopy and related results in other answers here.

share|cite|improve this answer
Interesting! Do you happen to know why this is called 'telescoping'? – drhab Jul 3 '14 at 19:58
@drhab The cancellation is analogous to collapsing the sections of a telescope down into a single section. As displayed above, when you rebracket the summands (sections) $\,f(i) = g(i) - g(i\!-\!1),$ the entire sum collapses to $\ 0 + \cdots + 0 + g(n) = g(n),\,$ the last "section". $\ \ $ – Bill Dubuque Jul 3 '14 at 20:16

I see lot's of good math here, but your repeated question of "Why are we adding (k+1)(k+2)?" leads me to believe you're looking for something a little different. (Of course, I could be completely wrong. If so, then pardon my rambing here.)

Try this:

Induction proofs often start with a formula that looks like

EQ $1.$ $\ $ term$_1$ + term$_2$ + ... + term$_n$ = $f(n)$

where $f(n)$ is a relatively simple function of $n$. We prove true for $1$, which is usually trivial, then assume it holds for $k$, so we have

EQ $2.$ $\ $ term$_1$ + term$_2$ + ... + term$_k$ = $f(k)$

Then we're supposed to prove it's true for $k+1$. OK, here's what no one is explicitly telling you. When you want to prove it's true for $k+1$, we take advantage of the fact that we know what those terms that make up the left-hand side of EQ $1$ look like, and that if we add the $k+1$ term to both sides of EQ $2$, then the new left-hand side looks like the first half of what we're trying to prove:

$\ $ term$_1$ + term$_2$ + ... + term$_k$ + term$_{k+1}$ = $f(k)$ + term$_{k+1}$

So, if we can do the algebra and rearrange that right hand side so it looks like $f(k+1)$ is supposed to , then we've got our proof. Here's where you plug $k+1$ into $f()$, to see what you're trying to manipulate that right-hand side to look like. If we can use algebra (or some other method) to show that

$f(k)$ + term$_{k+1} = f(k+1)$

then we've shown that adding the $k+1$ term makes the sequence equal to $f(k+1)$, which is what it means for the formula to be "true" for $k+1$.

Once you understand this, then the real trick to any of these is finding the right cool algebra trick to rearrange $f(k)$ + term$_{k+1}$ to get what you want, which is why so many of the answers you're getting focus on that aspect. It's the common question implied when someone says they don't get a particular induction proof.

(Note: This is a simple model with additive terms. Sometimes the pattern on the left is more complex and you have to figure out what you have to do to both sides to get from the $k$ to the $k+1$ case.)

(Also Note: This is just a verbose and somewhat less generalized case of what some of the above answers already say. However, I find that often students don't translate the symbology as readily as we "experts" (heh), especially in the early stages of learning a new concept, so I thought one more exposition from another tact might help.)

share|cite|improve this answer

You suppose that your formula is true for $k$ and prove that it is true for $(k+1)$. That is, suppose that: $$ 1\times2+\cdots+k\times(k+1)=\dfrac{k(k+1)(k+2)}{3}, $$ and prove for $(k+1)$: $$ 1\times2+\cdots+k\times(k+1)+(k+1)\times(k+2)=\dfrac{(k+1)(k+2)(k+3)}{3}. $$

Note that by assumption you have:

$$ 1\times2+\cdots+k\times(k+1)=\dfrac{k(k+1)(k+2)}{3}. $$

Hence: $$ \begin{equation} \begin{split} 1\times2+\cdots+(k+1)\times(k+2)&=\dfrac{k(k+1)(k+2)}{3}+(k+1)\times(k+2)\\&=\dfrac{k\color{red}{(k+1)(k+2)}+3\color{red}{(k+1)(k+2)}}{3}\\&=\dfrac{\color{red}{(k+1)(k+2)}(k+3)}{3}.\mathrm{QED} \end{split} \end{equation} $$

share|cite|improve this answer

Finishing up the induction proof,\begin{align} 1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + (k+1)(k+2) &= [1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + k(k+1)]+(k+1)(k+2) \\ &= \frac{k(k+1)(k+2)}{3}+(k+1)(k+2) \\ &= \frac{k(k+1)(k+2)}{3}+\frac{3(k+1)(k+2)}{3} \\ &= \frac{k(k+1)(k+2)+3(k+1)(k+2)}{3} \\ &= \frac{(k+1)(k+2)(k+3)}{3} \end{align} where the last equality is by hypothesis. Now simplify the last expression accordingly and you will get your result.

share|cite|improve this answer
Why are we adding (k+1)(k+2)? – user3718584 Jul 3 '14 at 18:49
If you combine the last equality in my answer, you'll actually factor out $(k+1)(k+2)$ as it appears on both terms. You will get $$\frac{(k+3)[(k+1)(k+2)]}{3}$$ which is your desired result. – Cookie Jul 3 '14 at 18:51

Based on hypothesis: $$\sum_{i=1}^{k}i\left(i+1\right)=\frac{1}{3}k\left(k+1\right)\left(k+2\right)$$ it must be shown that: $$\sum_{i=1}^{k+1}i\left(i+1\right)=\frac{1}{3}\left(k+1\right)\left(k+2\right)\left(k+3\right)$$ Here $k+1$ is 'plugged in' for $k$.

That is the inductionstep. How to do that? Well, $$\sum_{i=1}^{k+1}i\left(i+1\right)=\sum_{i=1}^{k}i\left(i+1\right)+\left(k+1\right)\left(k+2\right)=\frac{1}{3}k\left(k+1\right)\left(k+2\right)+\left(k+1\right)\left(k+2\right)$$ (in the last equality you are using the hypothesis)

So if you can show that the RHS of this indeed equals: $$\frac{1}{3}\left(k+1\right)\left(k+2\right)\left(k+3\right)$$ then you are ready. Working out the RHS comes to adding $(k+1)(k+2)$ to $\frac{1}{3}k\left(k+1\right)\left(k+2\right)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.