Understanding inductive proof of $\sum_{{m=k}}^{N} {m\choose k} = {N+1\choose K+1}$

Question

I am an engineering student and am trying to prove the following combinatorics identity in math:

$$\sum_{{m=k}}^{N} C(m,k) = C(N+1, K+1)$$

It was suggested to me to use Proof By Induction so I tried to do this problem.

• Step 1: Show this identity is true for a specific choice of $N=K$

LHS:

$$\sum_{{m=K}}^{N} C(m,k) = \sum_{{m=k}}^{k} C(m,k) = C(k,k) = 1$$

RHS: $$C(N+1, k+1)$$ $$\text{Substitute } N=k $$ $$C(k+1, k+1) = C(K+1, K+1) = 1$$

• Step 2: Assume this identity is true for any $N$

• Step 3: Prove this identity is true for $N=N+1$:

LHS: $$\sum_{{m=k}}^{N} C(m,k)$$ $$\text{Substitute } N=N+1 $$ $$ \sum_{{m=k}}^{N+1} C(m,k)$$

RHS:

$$C(N+1, K+1)$$ $$\text{Substitute } N=N+1 $$ $$C(N+1+1, k+1) = C(N+2, K+1)$$

Now, prove that LHS=RHS, i.e. $$\sum_{{m=k}}^{N+1} C(m,k) = C(N+1+1, k+1) = C(N+2, K+1)$$

We start by showing:

$$\sum_{{m=k}}^{N+1} C(m,K) = \color{red}{\sum_{{m=k}}^{N} C(m,k)} + \sum_{{m=N+1}}^{N+1} C(m,k) = \color{red}{C(N+1, k+1)} + C(N+1, k)$$

Note that $\color{red}{\sum_{{m=k}}^{N} C(m,k)} = \color{red}{C(N+1, k+1)}$ is assumed to be true because of Step 2.

Next, we continue the proof to show: $${C(N+1, k+1)} + C(N+1, k)$$

$$ = \frac{(N+1)!}{(K+1)!(N-K)!} + \frac{(N+1)!}{(K)!(N+1-K)!}$$

Using the idea that in general, $n! = n \cdot (n-1)!$, we write:

$$\frac{(N+1)!}{(K+1)!K!(N-k)!} + \frac{(N+1)!}{K!(N-k)!(N+1-k)}$$

Looking for common terms, we write:

$$\frac{(N+1)!}{K!(N-k)!} \left[ \frac{1}{(K+1)} + \frac{1}{(N+1-k)} \right]$$ $$= \frac{(N+1)!}{K!(N-K)!} \left[ \frac{N+1-K + K+1}{(K+1)(N+1-k)} \right]$$ $$=\frac{(N+1)!}{K!(N-K)!} \left[ \frac{N+2}{(K+1)(N+1-k)} \right]$$

Now, using the following identities based on $n! = n \cdot (n-1)!$ :

$$(N+1)! (N+2) = (N+2)!$$ $$K!(K+1) = (K+1)!$$ $$(N-k)!(N+1-K) = (N+1-k)!$$

We can now write:

$$\frac{(N+1)!}{K!(N-K)!} \left[ \frac{N+2}{(K+1)(N+1-k)} \right] = \frac{(N+2)!}{(K+1)!(N+1-k)!} = C(N+2, K+1)$$

And this is exactly what we were trying to show in Step 3:

$$\text{Original Identity: } $$ $$\sum_{{m=k}}^{N} C(m,k) = C(N+1, k+1)$$ $$\text{Substitute } N=N+1 $$ $$\sum_{{m=k}}^{N+1} C(m,k) = C(N+1+1, k+1) = C(N+2, K+1)$$

Thus, this concludes the proof. Note that Step 3 was not possible to prove without the assumption in Step 2 being a valid assumption (i.e. Step 3 serves to validate the assumption taken in Step 2).

Now, here is where my question begins: My prof told me that my work looks correct (the prof didn't do a full review, but looked at it quickly), but I personally don't agree with this proof. In general, I am confused how mathematical induction proofs work in general.

Here is my attempt to justify this proof:

• In Step 1, we proved that this identity true for a specific choice of $N=K$, e.g. $N=1$
• In Step 2, we assumed that this identity is true for any $N$. The way I see it as that suppose we took the case of $N=5$, we use $N=1$ as stepping stones to reach $N=5$
• But I am confused about Step 3. It seems the main argument is that Step 3 can only be true if our assumption about Step 2 is true.

But isn't the $N+1$ case contained within the $N$ case? If we assume this identity is true for the general $N$ case (as we did in Step 2), are we not forced to accept it must also be true for the $N+1$ case? When we say any $N$, isn't $N+1$ also "any $N$"? Is $N+1$ a general case? Can someone please help me understand how the argument of mathematical induction serves to prove this identity?

Answer 1

Can someone please help me understand how the argument of mathematical induction serves to prove this identity?

There is no such thing as a step 2 separate from step 3. In ordinary induction, the principle is that a statement is true, i.e. holds, if 1) it is true for the base case (or, base cases, plural, depending on the underlying inductive structure: anyway, let's stick to $\mathbb{N}$ here); and, 2) it is true for the successor case, i.e., if it is true for any $n$, then it is true for $n+1$.

More formally, for $P$ any property on the natural numbers, the induction principle looks like:

$$ (P(0) \land (\forall m \in \mathbb{N}, P(m) \to P(m+1))) \to \forall n \in \mathbb{N}, P(n) $$

so, it is sufficient to prove $P(0)$, the base case, and $\forall m \in \mathbb{N}, P(m) \to P(m+1)$, the successor case, then Modus Ponens from the induction principle gives $\forall n \in \mathbb{N}, P(n)$, though drawing this last conclusion is usually left implicit in informal argument under the heading of "proof by induction".

Answer 2

Seems to me that your confusion comes from the ideia of the induction itself, and not from this particular proof. Hence, I wil try to give you a insight on what is going on in the induction process.

The first step in a proof by induction is to prove the base case, for example, the case $n=1$. Then, as you pointed, our next step is to suppose that our statment is true for some $n \in \mathbb{N}$, and using this assumption we prove that it is true for $n+1$, in other words it is possible to see this process as a kind of "recursion": We know that it is true for $n=1$, and if the statment is true for $n\in\mathbb{N}$ it is true for $n+1$, then it is true for $n=2$, the same reasoning holds for $n=2$, thus it is true for $n=3$, and so on. Hence, it is true for every $n\in \mathbb{N}$.

I hope it helps. If you wanna learn induction in the correct way, I recommend you to take a look at the construction of natural numbers, where induction arises from one of the Peano Axioms.

As pointed by @Julio Di Egidio, this is not the right way to understand induction, but I believe that gives an insight on whats happening, specially for beginners. Once you have this initial intuition, it is important to understand the formal construction of natural number by Peano Axioms.