Chapter 9.5, Problem 59E

To determine

To expand: The expression ${\left(3\sqrt{t}+\sqrt[4]{t}\right)}^4$ by using the Binomial Theorem.

Answer to Problem 59E

The expansion of the expression ${\left(3\sqrt{t}+\sqrt[4]{t}\right)}^4$ is $81t^2+108t^{\frac{7}{4}}+54t^{\frac{3}{2}}+12t^{\frac{5}{4}}+t$ .

Explanation of Solution

Given information: The expression is ${\left(3\sqrt{t}+\sqrt[4]{t}\right)}^4$ .

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, $n=4$ .

First write out the Pascal’s triangle such that the row that begins with 1, 4.

  $\begin{array}{ccccccccc}& & & & 1& & & & \\ & & & 1& & 1& & & \\ & & 1& & 2& & 1& & \\ & 1& & 3& & 3& & 1& \\ 1& & 4& & 6& & 4& & 1\end{array}$

From the Pascal’s triangle, it is observed that the binomial coefficients of the expression are 1, 4, 6, 4 and 1.

Expand the expression ${\left(3\sqrt{t}+\sqrt[4]{t}\right)}^4$ by using formula in Binomial theorem as follows.

  $\begin{array}{c}{\left(3\sqrt{t}+\sqrt[4]{t}\right)}^4=\left(1\right){\left(3\sqrt{t}\right)}^4+4{\left(3\sqrt{t}\right)}^{4-1}\left(\sqrt[4]{t}\right)+6{\left(3\sqrt{t}\right)}^{4-2}{\left(\sqrt[4]{t}\right)}^2+4{\left(3\sqrt{t}\right)}^{4-3}{\left(\sqrt[4]{t}\right)}^3+\left(1\right){\left(\sqrt[4]{t}\right)}^4\\ =81t^2+4\left(27t^{\frac{3}{2}}\right)t^{\frac{1}{4}}+54t\left(t^{\frac{1}{2}}\right)+12t^{\frac{1}{2}}\left(t^{\frac{3}{4}}\right)+t\\ =81t^2+108t^{\frac{7}{4}}+54t^{\frac{3}{2}}+12t^{\frac{5}{4}}+t\end{array}$

Thus, the expansion of the expression ${\left(3\sqrt{t}+\sqrt[4]{t}\right)}^4$ is $81t^2+108t^{\frac{7}{4}}+54t^{\frac{3}{2}}+12t^{\frac{5}{4}}+t$ .