Proof of $(a+b)^{n+1}$

I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.

I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.

I however have trouble with the second one, I don’t know what to start exactly. Any help/hints?

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Note that $\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$, so


I think you need to understand the mathematical induction well.

You want to show the binomial theorem $(a+b)^n = \sum_{i = 0}^{n} \binom{n}{i}a^ib^{n-i}$, right?

To do this, You need to assume that this statement is true when $n = N$, then prove for the $n = N+1$ case with your assumption.

Then, if this is true for $n = 1$, then it is true for $n =2$, and so on for $n = 3,4, \cdots$ . So, for every natural number, the statement is true.

So, you should assume that $(a+b)^n = \sum\binom{n}{i}a^ib^{n-i}$, and prove $(a+b)^{n+1} = \sum\binom{n+1}{i}a^ib^{n+1-i}$.

I will show you a little example about induction; Prove $2^n > n$ for every natural number.

First, check the initial case; $2^1 = 2 >1$.

Now, assume that the statement is true when $n=N$, i.e. $2^N > N$.
Note that this is not the end of the proof; we did nothing yet. This is just an assumption. With this assumption, we will see what could be happened.

$2^{N+1} = 2^N \times 2 = 2^N + 2^N > 2N \ge N + 1$. (By our assumption)

This is $n = N+1$ case, so we proved the statement. Now the statement is true for $n = 2$, $n=3$, $\cdots$.