The binomial theorem states that for any real numbers a and b, $$( a + b ) ^ { n } = \sum _ { k = 0 } ^ { n } \left( \begin{array} { l }
n \
k
\end{array} \right) a ^ { n - k } b ^ { k } \quad \text { for any integer } n \geq 0$$
Use this theorem to show that for any integer $$n \geq 0 , \sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } \left( \begin{array} { l }
n \
k
\end{array} \right) 3 ^ { n - k } 2 ^ { k } = 1$$

Question

The binomial theorem states that for any real numbers a and b, $$( a + b ) ^ { n } = \sum _ { k = 0 } ^ { n } \left( \begin{array} { l } 
n \
k
\end{array} \right) a ^ { n - k } b ^ { k } \quad \text { for any integer } n \geq 0$$ 
Use this theorem to show that for any integer $$n \geq 0 , \sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } \left( \begin{array} { l } 
n \
k
\end{array} \right) 3 ^ { n - k } 2 ^ { k } = 1$$

Quizplus · Accepted Answer

The Answer of The binomial theorem states that for any real numbers a and b, $$( a + b ) ^ { n } = \sum _ { k = 0 } ^ { n } \left( \begin{array} { l } 
n \
k
\end{array} \right) a ^ { n - k } b ^ { k } \quad \text { for any integer } n \geq 0$$ 
Use this theorem to show that for any integer $$n \geq 0 , \sum _ { k = 0 } ^ { n } ( - 1 ) ^ { k } \left( \begin{array} { l } 
n \
k
\end{array} \right) 3 ^ { n - k } 2 ^ { k } = 1$$

The Binomial Theorem States That for Any Real Numbers a and B