Question

Expand ${\left(x+3\right)}^5$using Binomial Theorem

Short answer

The value of${\left(x+3\right)}^5=x^5+15x^4+90x^3+270x^2+405x+243$

Step-by-step solution

Step 1. Defination

Let$x$and$a$be real numbers 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

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

localid="1647227300943" $\Rightarrow {\left(x+3\right)}^5=\frac{5!}{0!\left(5-0\right)!}{x}^5+\frac{5!}{1!\left(5-1\right)!}\left(3\right)x^4+\frac{5!}{2!\left(5-2\right)!}\left(9\right)x^3+\frac{5!}{3!\left(5-3\right)!}\left(27\right)x^2+\frac{5!}{4!\left(5-4\right)!}\left(81\right)x^1+\frac{5!}{5!\left(5-5\right)!}\left(243\right)$

$\Rightarrow {\left(x+3\right)}^5=x^5+15x^4+90x^3+270x^2+405x+243$