Chapter 12.6, Problem 16E

To determine

To find: Expansion of ${\left(2+\frac{x}{2}\right)}^5$ using pascal’s triangle.

Answer to Problem 16E

  $\Rightarrow {\left(2+\frac{x}{2}\right)}^5=32+40x+20x^2+5x^3+\frac{5x^4}{8}+\frac{x^5}{32}$

Explanation of Solution

Given information:

  ${\left(2+\frac{x}{2}\right)}^5$

Calculation:

The Pascal’s Triangle is:

  $\begin{array}{l}1\\ 1-1\\ 1-2-1\\ 1-3-3-1\\ 1-4-6-4-1\\ 1-5-10-10-5-1\end{array}$

The triangle can be used to calculate the coefficients of the expansion of ${\left(a+b\right)}^n$ by taking the exponent $n$ and adding $1$ . The coefficients with correspond with the ${\left(n+1\right)}^{th}$ line of the triangle.

So, for ${\left(2+\frac{x}{2}\right)}^5$ the value of $n$ is $5$ . Therefore, the coefficients of the expansion will correspond to the $6^{th}$ line of the triangle.

The expansion of ${\left(a+b\right)}^n$ follows the rule, ${\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+\mathrm{.........}+C_{n-1}{a}^1{b}^{n-1}+C_n{a}^0{b}^n$

Where $C_0,C_1,C_2\mathrm{............}{C}_{n-1},C_n$ are the corresponding coefficients from the Pascal’s Triangle.

Therefore, here the Coefficients are $1-5-10-10-5-1$ .

  $\Rightarrow {\left(2+\frac{x}{2}\right)}^5=1{\left(2\right)}^5{\left(\frac{x}{2}\right)}^0+5{\left(2\right)}^4{\left(\frac{x}{2}\right)}^1+10{\left(2\right)}^3{\left(\frac{x}{2}\right)}^2+10{\left(2\right)}^2{\left(\frac{x}{2}\right)}^3+5{\left(2\right)}^1{\left(\frac{x}{2}\right)}^4+1{\left(2\right)}^0{\left(\frac{x}{2}\right)}^5$

  $\Rightarrow {\left(2+\frac{x}{2}\right)}^5=32+5{\left(2\right)}^4\left(\frac{x}{2}\right)+10{\left(2\right)}^3\left(\frac{x^2}{2^2}\right)+10\left(2^2\right)\left(\frac{x^3}{2^3}\right)+5\left(2\right)\left(\frac{x^4}{2^4}\right)+\left(\frac{x^5}{2^5}\right)$

  $\Rightarrow {\left(2+\frac{x}{2}\right)}^5=32+40x+20x^2+5x^3+\frac{5x^4}{8}+\frac{x^5}{32}$