Chapter 11.7, Problem 5WE

To determine

To calculate:

Expansion and simplification of expression

  ${\left(a+1\right)}^8$

Answer to Problem 5WE

The simplified expression is $a^8+8a^7+28a^6+56a^5+70a^4+56a^3+28a^2+8a+1$

Explanation of Solution

Given information:

The given expression is,

  ${\left(a+1\right)}^8$

Formula:

The binomial theorem is given as,

  ${\left(a+b\right)}^n={\displaystyle {\sum }_{i=0}^n\left(\right)}{a}^{\left(n-1\right)}{b}^i$

Calculation:

The given expression is,

  ${\left(a+1\right)}^8$

Here, consider

  $\begin{array}{l}a=a\\ b=1\end{array}$

  $={\displaystyle {\sum }_i^8=0\left(\right)}{a}^{\left(8-i\right)\cdot 1^i}$

Expanding the above expression using binomial theorem,

  $\begin{array}{l}=\frac{8!}{0!\left(8-0\right)}{\left(a\right)}^8\cdot 1^0+\frac{8!}{1!\left(8-1\right)}{\left(a\right)}^7\cdot 1^1+\frac{8!}{2!\left(8-2\right)}{\left(a\right)}^6\cdot 1^2+\frac{8!}{3!\left(8-3\right)}{\left(a\right)}^5\cdot 1^3\\ \frac{8!}{4!\left(8-4\right)}{\left(a\right)}^4\cdot 1^4+\frac{5!}{5!\left(8-5\right)}{\left(a\right)}^3\cdot 1^5+\frac{5!}{6!\left(8-6\right)}{\left(a\right)}^2\cdot 1^6+\frac{5!}{7!\left(8-7\right)}{\left(a\right)}^1\cdot 1^7+\frac{5!}{8!\left(8-8\right)}{\left(a\right)}^0\cdot 1^8\end{array}$

By simplifying the above factors the expression obtained is,

  $a^8+8a^7+28a^6+56a^5+70a^4+56a^3+28a^2+8a+1$