Chapter 12, Problem 13QC

To determine

To expand : The Equation ${\left(\frac{a}{2}+2\right)}^5$ using binomial

Answer to Problem 13QC

  $\frac{a^5}{32}+\frac{5}{3}{a}^4+5a^3+20a^2+40a+32$

Explanation of Solution

Given information :

The given information is, a binomial at ${\left(\frac{a}{2}+2\right)}^5$ .

Formula Used:

  ${(a+b)}^w={\displaystyle \sum _{r=0}^w{}^n{c}_r{a}^{n-r}{b}^r}$

Calculation : $\begin{array}{l}{\left(\frac{a}{2}+2\right)}^5={\displaystyle \sum _{r=0}^5{}^5{c}_r{\left(\frac{a}{2}\right)}^{5-r}\cdot 2^r}\\ ={}^5{c}_0{\left(\frac{a}{2}\right)}^{5-0}\cdot 2^0+{}^5{c}_1{\left(\frac{a}{2}\right)}^4\cdot 2^1+{}^5{c}_2{\left(\frac{a}{2}\right)}^3\cdot 2^2+{}^5{c}_3{\left(\frac{a}{2}\right)}^2\cdot 2^3+{}^5{c}_4{\left(\frac{a}{2}\right)}^1\cdot 2^4+{}^5{c}_5{\left(\frac{a}{2}\right)}^0\cdot 2^5\\ ={\left(\frac{a}{2}\right)}^5+5\left(\frac{a^4}{16}\right)\cdot 2+10\frac{a^3}{8}\cdot 4+10\frac{a^2}{4}\cdot 8+5\frac{a}{2}\cdot 16+32\\ =\frac{a^5}{32}+\frac{5}{8}{a}^4+5a^3+20a^2+40a+32\\ {\left(\frac{a}{2}+2\right)}^5=\frac{a^5}{32}+\frac{8}{5}{a}^4+5a^3+20a^2+40a+32\end{array}$