Question

Expand ${\left(ax+by\right)}^5$using Binomial Theorem

Short answer

The value of${\left(ax+by\right)}^5=a^5{x}^5+5a^{4}b{x}^{4}y+10a^3{b}^2{x}^3{y}^2+10a^2{b}^3{x}^2{y}^3+5a{b}^{4}x{y}^4+b^5{y}^5$

Step-by-step solution

Step 1. Defination

Let $x$and $a$be real number for any positive integer$n$, we have${\left(x+a\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\cdots \cdots +\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+\cdots \cdots +\left(\begin{array}{c}n\\ n\end{array}\right)a^n=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)x^{n-j}{a}^j$

Step 2. Explanation

$\Rightarrow {\left(ax+by\right)}^5=\left(\begin{array}{c}5\\ 0\end{array}\right){\left(by\right)}^0{\left(ax\right)}^5+\left(\begin{array}{c}5\\ 1\end{array}\right){\left(by\right)}^1{\left(ax\right)}^4+\left(\begin{array}{c}5\\ 2\end{array}\right){\left(by\right)}^2{\left(ax\right)}^3+\left(\begin{array}{c}5\\ 3\end{array}\right){\left(by\right)}^3{\left(ax\right)}^2+\left(\begin{array}{c}5\\ 4\end{array}\right){\left(by\right)}^4{\left(ax\right)}^1+\left(\begin{array}{c}5\\ 5\end{array}\right){\left(by\right)}^5{\left(ax\right)}^0$

$\Rightarrow {\left(ax+by\right)}^5=a^5{x}^5+5a^{4}b{x}^{4}y+10a^3{b}^2{x}^3{y}^2+10a^2{b}^3{x}^2{y}^3+5a{b}^{4}x{y}^4+b^5{y}^5$