In my probability textbook, I saw this summation, $\sum_{n=0}^{N}$$N\choose{n}$$s^n$$=(1+s)^N$ but I have no idea why it stands, could someone please show me the steps in between? (Or link me something that is similar). Much appreciated!
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It should be $(1+s)^N$. – Wuestenfux Sep 18 '19 at 11:44
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It is a finite sum. Convergence makes sense only for infinite sums. – Somos Sep 18 '19 at 12:23
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Isn't that the definition of "binomial coefficients"? – mr_e_man Sep 18 '19 at 23:14
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By the Binomial Theorem $$(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$$
b00n heT
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If you are not familiar with the binomial theorem, induction is one way to see this.
($N=1$) $$ \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} $$
luxerhia
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