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# Binomial Theorem

$(p + q)^n = \sum\limits_{k=0}^n {n \choose k} p^k q^{n-k}$

$\sum\limits_{j=1}^{\infty} a^j = {a \over {1-a}}, |a| < 1$

$S_N = \sum\limits_{j=1}^N a^j$

$S_N = a + a^2 + \cdots + a^N$ — ①

$aS_N = a^2 + \cdots a^{N+1}$ — ②

If we subtract ② from ①, we get

$S_N = {{a - a^{n+1}} \over {1-a}}$

$\sum\limits_{k=1}^{\infty} k a^k = {a \over (1-a)^2}$

$\lim\limits_{n \to \infty} ({1 + x \over n})^n = e^x$

Number of OutcomesWith ReplacementWithout Replacements
2Binomial (different when $\text{until}^\text{*}$...)Hypergeometric
$\geq$ 3MultinomialMultivariate Hypergeometric

## $\text{until}^\text{*}$​

• $1^{\text{st}}$ success → geometric
• $r^{\text{th}}$ success → negative binomial

In Probability,

$(p+q)^n = \sum\limits_{k=0}^n {n \choose k} p^k q^{n-k}$

## Proof​

### Base case​

Let $n=1$. Then

$\sum\limits_{k=0}^1 {1 \choose k }p^k q^{n-k} = {1 \choose 0} p^0 q^1 + {1 \choose 1} p^1 q^0$

### Induction Hypothesis​

Assume that $(p+q)^m = \sum\limits_{k=0}^m {1 \choose m }p^k q^{m-k}$. Let $n = m+1$. Then

$(p+q)^{m+1} = (p+q) (p+q)^m = (p+q) \sum\limits_{k=0}^m {1 \choose m }p^k q^{m-k}$ $p \sum\limits_{k=0}^m {m \choose k}p^k q^{m-k} + q \sum\limits_{k=0}^m {m \choose k}p^k q^{m-k}$ $\sum\limits_{k=0}^m {m \choose k}p^{k+1} q^{m-k} + \sum\limits_{k=0}^m {m \choose k}p^k q^{m-k+1}$

Let $j=k+1$ and $k=j-1$ for the first and $j=k$ for the second sum. Then,

$\sum\limits_{j=1}^{j=m+1} {m \choose {j-1}}p^{j} q^{m + 1 - j} + \sum\limits_{j=0}^{j=m} {m \choose j}p^j q^{m-j+1}$