Section 12.2 Binomial SumsBinomial SeriesTrinomial Series
The binomial series is also foundational. It is technically not a series since the sum_if finite but we won’t bother with that for now. It is given by
\begin{gather*}
B = \sum_{k=0}^{n} {\binom{n}{k} a^k b^{n-k}}
\end{gather*}
Theorem 12.2.1 Binomial Theorem
For \(n \in \mathbb{N} \text{,}\) \(\displaystyle {(a+b)^n = \sum_{k=0}^{n} {\binom{n}{k} a^k b^{n-k}}}\)Proof
By induction:
Basic Step: n = 1 is trivial
Inductive Step: Assume the statement is true as given for some \(n \ge 1\text{.}\) Show \((a+b)^{n+1} = \sum_{k=0}^{n+1} {\binom{n+1}{k} a^k b^{n+1-k}}\)
\begin{align*}
(a+b)^{n+1} & = (a+b)(a+b)^n\\
& = (a+b)\sum_{k=0}^{n} {\binom{n}{k} a^k b^{n-k}}\\
& = \sum_{k=0}^n \binom{n}{k} a^{k+1} b^{n-k} + \sum_{k=0}^n \binom{n}{k} a^k b^{n-k+1}\\
& = \sum_{k=0}^{n-1} \binom{n}{k} a^{k+1} b^{n-k} + a^{n+1} + b^{n+1} + \sum_{k=1}^n \binom{n}{k} a^k b^{n-k+1}\\
& = \sum_{j=1}^n \binom{n}{j-1} a^j b^{n-(j-1)} + a^{n+1} + b^{n+1} + \sum_{k=1}^n \binom{n}{k} a^k b^{n+1-k}\\
& = b^{n+1} + \sum_{k=1}^n \left [ \binom{n}{k-1} + \binom{n}{k} \right ] a^k b^{n+1-k} + a^{n+1}\\
& = b^{n+1} + \sum_{k=1}^n \binom{n+1}{k} a^k b^{n+1-k} + a^{n+1}\\
& = \sum_{k=0}^{n+1} \binom{n+1 }{k} a^k b^{n+1-k}
\end{align*}
Consider \(B(a,b) = \sum_{k=0}^{n} {\binom{n}{k} a^k b^{n-k}}\text{.}\) This finite sum_is known as the Binomial Series.
Show that \(B(a,b) = (a+b)^n\)
Show that \(B(1,1) = 2^n\)
Show that \(B(-1,1) = 0\)
Show that \(B(p,1-p) = 1\)
Easily, \(B(x,1) = \sum_{k=0}^{n} {\binom{n}{k} a^k}\)
\begin{gather*}
(a+b+c)^n = \sum_{k_1+k_2+k_3=n}^{} {\binom{n}{k_1,k_2,k_3} a^{k_1} b^{k_2} c^{k_3}}
\end{gather*}
where \(\binom{n}{k_1,k_2,k_3} = \frac{n!}{k_1!k_2!k_3!}\text{.}\) This can be generalized to any number of terms to give what is know as a multinomial series.