The binomial theorem states that for any real numbers a and b, $( a + b ) ^ { n } = \sum _ { k = 0 } ^ { n } \left( \begin{array} { l } n \ k \end{array} \right) a ^ { n - k } b ^ { k }$ for any integer $n \geq 0$.
Use this theorem to compute $( 2 x - 1 ) ^ { 5 }$

Question

The binomial theorem states that for any real numbers a and b, $( a + b ) ^ { n } = \sum _ { k = 0 } ^ { n } \left( \begin{array} { l } n \ k \end{array} \right) a ^ { n - k } b ^ { k }$ for any integer $n \geq 0$. 
Use this theorem to compute $( 2 x - 1 ) ^ { 5 }$

Quizplus · Accepted Answer

The Answer is Apply the binomial theorem with a = 2x, b = −1, and n = 5.

\begin{aligned}( 2 x - 1 ) ^ { 5 } & = \sum _ { k = 0 } ^ { 5 } \left( \begin{array} { l } 5 \k\end{array} ight) a ^ { 5 - k } b ^ { k } \& = \left( \begin{array} { l } 5 \0\end{array} ight) a ^ { 5 - 0 } b ^ { 0 } + \left( \begin{array} { l } 5 \1\end{array} ight) a ^ { 5 - 1 } b ^ { 1 } + \left( \begin{array} { l } 5 \2\end{array} ight) a ^ { 5 - 2 } b ^ { 2 } + \left( \begin{array} { l } 5 \3\end{array} ight) a ^ { 5 - 3 } b ^ { 3 } + \left( \begin{array} { l } 5 \4\end{array} ight) a ^ { 5 - 4 } b ^ { 4 } + \left( \begin{array} { l } 5 \5\end{array} ight) a ^ { 5 - 5 } b ^ { 5 } \& = a ^ { 5 } + 5 a ^ { 4 } b + 10 a ^ { 3 } b ^ { 2 } + 10 a ^ { 2 } b ^ { 3 } + 5 a b ^ { 4 } + b ^ { 5 } \& = a ^ { 5 } + 5 a ^ { 4 } b + 10 a ^ { 3 } b ^ { 2 } + 10 a ^ { 2 } b ^ { 3 } + 5 a b ^ { 4 } + b ^ { 5 } \& = ( 2 x ) ^ { 5 } + 5 ( 2 x ) ^ { 4 } ( - 1 ) + 10 ( 2 x ) ^ { 3 } ( - 1 ) ^ { 2 } + 10 ( 2 x ) ^ { 2 } ( - 1 ) ^ { 3 } + 5 ( 2 x ) ( - 1 ) ^ { 4 } + ( - 1 ) ^ { 5 } \& = 32 x ^ { 5 } + 5 \cdot 16 x ^ { 4 } \cdot ( - 1 ) + 10 \cdot 8 x ^ { 3 } \cdot 1 + 10 \cdot 4 x ^ { 2 } \cdot ( - 1 ) + 5 ( 2 x ) \cdot 1 + ( - 1 ) \& = 32 x ^ { 5 } - 80 x ^ { 4 } + 80 x ^ { 3 } - 40 x ^ { 2 } + 10 x - 1\end{aligned}

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