23.4: Exercises
- Page ID
- 83520
Choose numbers \(x,y\) so that the equality in the Binomial Theorem becomes
\begin{equation*} \sum_{k=0}^n \binom{n}{k}\, 2^k = 3^n \text{.} \end{equation*}
- Choose numbers \(x,y\) so that the equality in the Binomial Theorem becomes
\begin{equation*} \binom{n}{0} \;\;-\;\; \binom{n}{1} \;\;+\;\; \binom{n}{2} \;\;-\;\; \binom{n}{3} \;\;+\;\; \cdots \;\;+\;\; (-1)^n \binom{n}{n} \;\;=\;\; 0\text{.} \end{equation*}
- The equality from Task a can be rearranged to yield
\begin{gather*} \binom{n}{0} \;\;+\;\; \binom{n}{2} \;\;+\;\; \binom{n}{4} \;\;+\;\; \cdots \;\;+\;\; \binom{n}{m_1}\\ \;\;=\;\; \binom{n}{1} \;\;+\;\; \binom{n}{3} \;\;+\;\; \binom{n}{5} \;\;+\;\; \cdots \;\;+\;\; \binom{n}{m_2}\text{,} \end{gather*}
where
\begin{align*} m_1 & = \begin{cases} n, & n\text{ even}, \\ n-1, & n\text{ odd}, \\ \end{cases} & m_2 & = \begin{cases} n-1, & n\text{ even}, \\ n, & n\text{ odd}. \\ \end{cases} \end{align*}
What does this rearranged formula tell you about the subsets of a set of size \(n\text{?}\)
- Hint.
-
What is the sum on the left counting? What is the sum on the right counting?