Chapter 12.5, Problem 35AYU

In Problems 29-42, use the Binomial Theorem to find the indicated coefficient or term.

The 5th term in the expansion of ${\left(x+3\right)}^7$

To determine

To find: The 5^th term expansion of the given binomial expression ${\left(x+3\right)}^7$ using Binomial Theorem.

Answer to Problem 35AYU

Solution:

The 5^th term in the expansion is $2835x^3$.

Explanation of Solution

Given:

Expression is given as ${\left(x+3\right)}^7$.

Formula used:

The Binomial Theorem:

Let $x$ and $a$ be real numbers. For any positive integer $n$, we have

$\begin{array}{l}{\left(x + a\right)}^n = \left(\begin{array}{c}n\\ 0\end{array}\right)x^n + \left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1} + \cdots + \left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j} + \cdots + \left(\begin{array}{c}n\\ n\end{array}\right)a^n\\ {\left(x + a\right)}^n = {\displaystyle \sum _{k = 0}^n\left(\begin{array}{c}n\\ k\end{array}\right)x^{n-j}{a}^j}\end{array}$

Here $a = 3$ and $n = 7$. Applying Binomial theorem till it reached 5^th term

$\begin{array}{l}{\left(x + 3\right)}^7 = \left(\begin{array}{c}7\\ 0\end{array}\right)x^7 + \left(\begin{array}{c}7\\ 1\end{array}\right)\left(3\right)x^6 + \left(\begin{array}{c}7\\ 2\end{array}\right){\left(3\right)}^2{x}^5 + \left(\begin{array}{c}7\\ 3\end{array}\right){\left(3\right)}^3{x}^4 + \left(\begin{array}{c}7\\ 4\end{array}\right){\left(3\right)}^4{x}^3 + \cdots \\ \left(\begin{array}{c}7\\ 4\end{array}\right){\left(3\right)}^4{x}^3 = 35 \times 81.x^3 = 2835x^3\end{array}$