Theorem 1.5

#Theorem

If $a$ and $b$ are real numbers and $n$ is a positive integer, then $(a + b)^{n} = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k}$

Proof: $(a + b)^{n} = (a + b) \dots (a + b) (n factors)$

In the expansion of $(a + b)^{n}$ , a term of the form $a^{n - k} b^{k}$ arises from choosing $b$ from $k$ factors and $a$ from the other $n - k$ factors. This can be done in $(\binom{n}{k})$ ways. Thus $a^{n - k} b^{k}$ appears $(\binom{n}{k})$ times in the expansion. It follows that:

\begin{aligned} (a + b)^{n} \\ = (\binom{n}{0}) a^{n - 1} b^{1} + (\binom{n}{2}) a^{n - 2} b^{2} + \dots + (\binom{n}{n - 1}) a^{1} b^{n - 1} + (\binom{n}{n}) a^{0} b^{n} \\ = \sum_{k = 0}^{n} (\binom{n}{k}) a^{n - k} b^{k} \end{aligned}

The numbers $(\binom{n}{k})$ are known as binomial coefficients because they appear in the expansion of $(a + b)^{n}$

Proof: (a+b)n=(a+b)…(a+b) (n factors)

Proof: $(a + b)^{n} = (a + b) \dots (a + b) (n factors)$