Chapter 11.7, Problem 1OE

To determine

To calculate:

The first four terms for the Pascal triangle row $8$ :

  ${\left(a+b\right)}^8$

Answer to Problem 1OE

The first four terms for the Pascal triangle row $8$ is

  ${\left(a+b\right)}^8=a^8+8a^{7}b+28a^6{b}^2+56a^5{b}^3+\mathrm{...}+1b^8$ .

Explanation of Solution

Given information:

  $\begin{array}{l}n=8\\ C_0=1\\ C_1=8\\ C_2=28\\ C_3=56\\ C_4=70\\ C_5=56\\ C_6=28\\ C_7=8\\ C_8=1\end{array}$

Formula used:

The binomial theorem using the Pascal’s triangle:

  $\begin{array}{l}{\left(a+b\right)}^n=a^n+n{a}^{n-1}b+\frac{n\left(n-1\right)}{2!}{a}^{n-2}{b}^2+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{a}^{n-3}{b}^3+\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}{a}^{n-4}{b}^4+\mathrm{...}+b^n\\ {\left(a+b\right)}^n=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n\end{array}$

Calculation:

The binomial theorem for the given Pascal’s triangle is derived as:

  ${\left(a+b\right)}^8=C_0{a}^n{b}^0+C_1{a}^{n-1}{b}^1+C_2{a}^{n-2}{b}^2+C_3{a}^{n-3}{b}^3+C_4{a}^{n-4}{b}^4+\mathrm{...}+C_n{a}^0{b}^n$

Put the given data:

  $\begin{array}{l}{\left(a+b\right)}^8=\left(1\right)a^8{b}^0+\left(8\right)a^{8-1}{b}^1+\left(28\right)a^{8-2}{b}^2+\left(56\right)a^{8-3}{b}^3+\mathrm{...}+1a^0{b}^8\\ {\left(a+b\right)}^8=a^8+8a^{7}b+28a^6{b}^2+56a^5{b}^3+\mathrm{...}+1b^8\left[\because x^0=1\right]\end{array}$