Chapter 8.4, Problem 78E

To determine

To expand: The Binomial ${\left(3x+4y\right)}^5$ by using the Binomial Theorem.

Answer to Problem 78E

The expansion of the expression ${\left(3x+4y\right)}^5$ is $243x^5+1620x^{4}y+4320x^3{y}^2+17280x^2{y}^3+3840x{y}^4+1024y^5$ .

Explanation of Solution

Given information: The expression is ${\left(3x+4y\right)}^5$ .

Theorem used: The Binomial Theorem: ${\left(x+y\right)}^n=x^n+n{x}^{n-1}y+\cdots {+}_n{C}_r{x}^{n-r}{y}^r+\cdots +nx{y}^{n-1}+y^n$ , the coefficient of $x^{n-r}{y}^r$ is ${}_n{C}_r=\frac{n!}{\left(n-r\right)!r!}$ .

Calculation:

Here, $n=5$ .

First write out the Pascal’s triangle such that the row that begins with 1, 5.

  $\begin{array}{ccccccccccc}& & & & & 1& & & & & \\ & & & & 1& & 1& & & & \\ & & & 1& & 2& & 1& & & \\ & & 1& & 3& & 3& & 1& & \\ & 1& & 4& & 6& & 4& & 1& \\ 1& & 5& & 10& & 10& & 5& & 1\end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 5, 10, 10, 5 and 1.

Expand the expression ${\left(3x+4y\right)}^5$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(3x+4y\right)}^5=\left(1\right){\left(3x\right)}^5{\left(4y\right)}^0+5{\left(3x\right)}^{5-1}\left(4y\right)+10{\left(3x\right)}^{5-2}{\left(4y\right)}^2+10{\left(3x\right)}^{5-3}{\left(4y\right)}^3+5{\left(3x\right)}^{5-4}{\left(4y\right)}^4+\left(1\right){\left(4y\right)}^5\\ =243x^5+20{\left(3x\right)}^{4}y+160{\left(3x\right)}^3{y}^2+640{\left(3x\right)}^2{y}^3+1280\left(3x\right)y^4+1024y^5\\ =243x^5+1620x^{4}y+4320x^3{y}^2+17280x^2{y}^3+3840x{y}^4+1024y^5\end{array}$

Thus, the expansion of the expression ${\left(2x+3y\right)}^5$ is $243x^5+1620x^{4}y+4320x^3{y}^2+17280x^2{y}^3+3840x{y}^4+1024y^5$ .