6.3a Binomial Theorem

Binomial Theorem

For all $n\in \mathbb{N}$:
$(a+b{)}^n={\sum }_{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})a^k{b}^{n-k}$

Proof

Take a look at the $n$-fold product $(a+b{)}^n=(a+b)(a+b)\dots (a+b)$, such that for $n=3$,
$(a+b{)}^3=(a^2+2ab+b^2)(a+b)=a^3+a^{2}b+2a^{2}b+2a{b}^2+a{b}^2+b^3=a^3+3a^{2}b+3a{b}^2+b^3,$
where each monomial can be expressed as some $c\cdot a^{n-k}{b}^k$ or vice-versa. In simple language, we are trying to find all the arrangements of $n-k$ factors of $a$ and $k$ factors of $b$. For $n=3$ and $k=1$, we can do $(b)(a)(a)$, $(a)(b)(a)$, and $(a)(a)(b)$, thereby producing $c=3$.