# Mathematical induction proof that $\sum\limits_{k=2}^{n-1} {k \choose 2} = {n \choose 3}$

I have a problem with a mathematical equation. I don’t find the given solution.

This is the equation: $\sum\limits_{k=2}^{n-1} {k \choose 2} = {n \choose 3}$

I should show with induction that the expression is correct for every $n >= 3$.

I started with the beginning of the induction. For $n = 3$.

$\sum\limits_{k=2}^{2} {2 \choose 2} = 1 = {3 \choose 3}$ That’s correct.

Now I don’t know how to dissolve that equation. The solution should be:

$\sum\limits_{k=2}^{n-1} {k \choose 2} + {n \choose 2} = {n \choose 3} + {n \choose 2} = {n + 1 \choose 3}$

Why ${n \choose 2}$? It would be awesome, if someone could tell me the steps or give some tips how I could reach that solution.

Edit: Well, at that point I am:

1. Beginning of the induction: $\sum\limits_{k=2}^{2} {2 \choose 2} = 1 = {3 \choose 3}$

2. Induction step n + 1: $\sum\limits_{k=2}^{n} {k \choose 2} + {n \choose 3}$ Thats my assumption: ${n \choose 3}$.

3. Change the first addend with the assumption, so I get ${n \choose 2} + {n \choose 3}$

4. Add it into the binomial coefficient formula: ${n! \choose 2!(n-2)!} + {n! \choose 3!(n-3)!}$

I have it like @david-mitra. Whats wrong with that way?

Thanks.

• Please inline your equations in the benefit of the users here. – user21436 Jan 16 '12 at 14:20
• Sorry, im new to the math stackexchange. What do you mean with inline? Do you mean MathJaX? Greetz. – user23053 Jan 16 '12 at 14:30
• Are you trying to show that $\sum\limits_{k=2}^{n-1}{k\over 2}={n\over3}$ for all $n\ge 3$? This isn't true... – David Mitra Jan 16 '12 at 14:36
• There is something drastically wrong on the left-hand side of your equation. Go back and look at the source of the problem. – ncmathsadist Jan 16 '12 at 14:59

We wish to prove: $$\sum_{k=2}^{n-1} {k\choose 2}= {n\choose3},\quad n\ge3.\tag{1}$$

If you want to prove that equation (1) holds for all $n\ge 3$ using induction:

1) Show that the equation is true for $n=3$:

$$\sum_{k=2}^{3-1} {k\choose 2}= {2\choose2}= {3\choose3}.\tag{1}$$

2) Assume that the equation is true for $n=m$: $$\color{maroon}{\sum_{k=2}^{m-1} {k\choose 2}}= \color{maroon}{m\choose3 }.$$

3) Now show that the equation must be true for $n=m+1$:

\eqalign{ \sum_{k=2}^{(m+1)-1} {k\choose 2} &=\sum_{k=2}^{m } {k\choose 2}\cr &=\sum_{k=2}^{m-1 } {k\choose2} +\color{darkgreen}{\sum_{k=m}^m {k\choose2}}\cr &=\biggl[\ \color{maroon}{ \sum_{k=2}^{m-1 } {k\choose 2}}\ \biggr]+\color{darkgreen}{m\choose2}\cr &= \color{maroon}{ m\choose 3} +{m\choose2}\cr &= {m!\over (m-3)! 3!}+ {m!\over (m-2)! 2!}\cr &={ m(m-1)(m-2)\over 3!} +{ m(m-1) \over 2!} \cr &= m(m-1)({m-2\over 6}+{3\over 6 })\cr &= { m(m-1)(m+1)\over 6} \cr &={m+1\choose 3}. }

• I don't think simple problem like this needs algebraic way, it can be shown with combinatorial way simply. – Saeed Jan 16 '12 at 23:16

Your equation is a special case of the identity $$\sum_{j=k}^{n-m}\binom{j}{k}\binom{n-j}{m}=\binom{n+1}{k+m+1}\tag{1}$$ with $m=0$.

Equation $(1)$ can also be shown by induction, but it also follows from the identity $$\frac{1}{(1-x)^{k+1}}=\sum_{n=k}^\infty\binom{n}{k}x^{n-k}\tag{2}$$ while considering $$\frac{1}{(1-x)^{k+1}}\frac{1}{(1-x)^{m+1}}=\frac{1}{(1-x)^{k+m+2}}\tag{3}$$

Here is how to solve your problem. You should know the principle of mathematical induction: Suppose you wish to prove that a proposition $P(n)$ is true for all integers $n \geq 3$, say.

If $P(3)$ is true (the basis step), and whenever $P(n)$ is true (the inductive step) it implies that $P(n+1)$ is true then $P(n)$ is true for all natural numbers $n \geq 3$.

This is the most basic form of induction that I learnt at high school, not even talking of things like the strong form of induction.

I will not solve your problem explicitly for then you will learn nothing. Instead, here is a guide.

Let $P(n)$ be the proposition that you wish to prove, namely

"For all integers $n \geq 3$, $\sum_{k=2}^{n-1} \binom{k}{2} = \frac{n(n-1)(n-2)}{6}$."

Now I like to think of induction in terms of dominoes. You have a countable number of dominoes stacked next to each other. You want to show that all the dominoes will fall. What do you do? Well you need to make sure that the first domino falls. This is the basis step in induction.

$\color{red}{\text{Step 1 (Basis step):}}$ for $n=3$, the left hand side is $1$ while the right hand side is $(3 \times 2 \times 1)/6 = 1$ so the base case holds.

Now the next step is to assume that given any other domino in your list, it does fall. This is the inductive step:

$\color{red}{\text{Step 2 (Inductive step):}}$ Assume that $P(i)$ holds for an integer $i \geq 1$, viz. it is indeed the case that for such an $i$,

$$\sum_{k=2}^{i-1} = \frac{i(i-1)(i-2)}{6}.$$

Now comes the hard part ($\color{red}{\text{Step 3}}$). You want to show that given a domino that falls, it does indeed hit the domino after it to make that fall too. In other words you want to show that $\color{blue}{\textit{if}}$ $P(i)$ is true $\color{blue}{\textit{then}}$ $P(i+1)$ is true. In other words, assuming that

$$\sum_{k=2}^{i-1} \binom{k}{2} = \frac{i(i-1)(i-2)}{6} \hspace{1in} (a)$$

is true somehow leads to the conclusion that

$$\sum_{k=2}^{i} \binom{k}{2}= \frac{(i+1)(i)(i-1)}{6}. \hspace{1in} (b)$$

Now we get to the heart of the problem. Why does one add for the matter $1/2$ or $1/3$ or whatever to both sides of $(a)$? First of all do you know what we will add to the left side? Well somehow from $(a)$ we want to get to $(b)$. Well if you look at the left hand sides of both $(a)$ and $(b)$, we get $(b)$ from $(a)$ by $\color{green}{\textit{adding}}$ $\binom{i}{2}$ to the left hand side of $(a)$. So if we add $\binom{i}{2}$ to $\color{green}{\textit{the right hand side}}$ of $(a)$ as well and we get $\color{green}{\textit{the right hand side}}$ of $(b)$ you have shown:

The assumption in $(a)$ being true has led to $(b)$ being true, viz. assuming that your proposition $P(n)$ being true for $n=i$ has led to $P(i+1)$ being true. Therefore by induction,

$P(n)$ is true for all positive integers $n$.

I leave you to check the arithmetic that adding $\binom{i}{2}$ to the right hand side of $(a)$ does indeed give the right hand side of $(b)$.

Regards.

HINT $\$ Prove by induction $\rm\displaystyle\ f(n)\: = \sum_{\ \: k\ =\ 2_{\phantom .}}^{n-1}\ g(k)\ \iff\ f(n+1) - f(n)\ =\ g(n),\quad f(3) = g(2)$
Then your exercise reduces to verifying the RHS equations (a rote polynomial computation), viz.

$$\rm\ {n+1\choose 3} - {n\choose 3}\ =\ {n\choose 2},\qquad {3\choose 3}\ =\ {2\choose 2}$$

For many further examples see my prior posts on telescopy.

I'll skip the start of proof as you said, I'll provide combinatorial proof (by induction), but For $n\ge 3$ for choosing $3$ item from $n+1$ item fix latest item $a_{n+1}$ then you have:

1. You selecting latest item with two items from first n items, which is ${n \choose 2}$
2. You selecting all three items from first $n$ items and as induction previous step is $\sum\limits_{k=2}^{n-1} {k \choose 2}$

So your total possible way (for $n+1$) is:

$$\sum\limits_{k=2}^{n-1} {k \choose 2} + {n \choose 2} = \sum\limits_{k=2}^{n} {k \choose 2} = {n+1 \choose 3}$$

• Also you can simply write $\sum\limits_{k=2}^{n-1} {k \choose 2}$ as ${n \choose 3}$ so total equation will be ${n \choose 3} + {n \choose 2} = {n+1 \choose 3}$ – Saeed Jan 16 '12 at 15:28
• In fact, $\sum_{k=0}^n\binom{k}{r}=\binom{n+1}{r+1}$ for all nonnegative integers $n,r$. Saeed's combinatorial argument can be modified to prove this equality too. – Batominovski Jun 21 '15 at 5:29