Could someone please show me the steps of this sum?

Question

In my probability textbook, I saw this summation, $\sum_{n=0}^{N}$$N\choose{n}$$s^n$$=(1+s)^N$ but I have no idea why it stands, could someone please show me the steps in between? (Or link me something that is similar). Much appreciated!

Answer 1

If you are not familiar with the binomial theorem, induction is one way to see this.

($N=1$) $$ \sum_{n=0}^{1} \binom{1}{n} s^n = 1+s=(1+s)^1$$

($N=m \to N=m+1$)

\begin{align*} (1+s)^{N+1} &= (1+s)(1+s)^N=(1+s)\sum_{n=0}^{N} \binom{N}{n} s^n \\&= \sum_{n=0}^{N} \binom{N}{n} s^n+\sum_{n=0}^{N} \binom{N}{n} s^{n+1} \\ &= 1+\sum_{n=1}^{N} \binom{N}{n} s^n + \sum_{n=1}^{N} \binom{N}{n-1} s^n + s^{N+1} \\ &= 1+\sum_{n=1}^{N} \left( \binom{N}{n} + \binom{N}{n-1} \right) s^n + s^{N+1} \\ &= 1+\sum_{n=1}^{N} \binom{N+1}{n} s^n + s^{N+1} \\& = \sum_{n=0}^{N+1} \binom{N+1}{n} s^n \end{align*}

Here I use the recursive formula $$ \binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1} $$

Answer 2

By the Binomial Theorem $$(a+b)^N=\sum_{n=0}^{N}{N\choose n}a^n\cdot b^{N-n}$$ choosing $a=s$ and $b=1$ leads to $$(1+s)^N=(s+1)^N=\sum_{n=0}^{N}{N\choose n}s^n\cdot 1^{N-n}=\sum_{n=0}^{N}{N\choose n}s^n$$