Answer:
[tex]$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$[/tex]
Step-by-step explanation:
[tex]$\left(\frac{1}{2}x+3y \right)^4=\left(\frac{x}{2}+3y \right)^4\\$[/tex]
Binomial Expansion Formula:
[tex]$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$[/tex], also [tex]$\binom{n}{k}=\frac{n!}{(n-k)!k!}$[/tex]
We have to solve [tex]$\left(\frac{x}{2} + 3 y\right)^{4}=\sum_{k=0}^{4} \binom{4}{k} \left(3 y\right)^{4-k} \left(\frac{x}{2}\right)^k$[/tex]
Now we should calculate for [tex]k=0, k=1, k=2, k=3 \text{ and } k =4;[/tex]
First, for [tex]k=0[/tex]
[tex]$\binom{4}{0} \left(3 y\right)^{4-0} \left(\frac{x}{2}\right)^{0}=\frac{4!}{(4-0)! 0!}\left(3 y\right)^{4} \left(\frac{x}{2}\right)^{0}=\frac{4!}{4!}(81y^4)\cdot 1 =81 y^{4}$[/tex]
It is the same procedure for the other:
For [tex]k=1[/tex]
[tex]$\binom{4}{1} \left(3 y\right)^{4-1} \left(\frac{x}{2}\right)^{1}=54 x y^{3}$[/tex]
For [tex]k=2[/tex]
[tex]$\binom{4}{2} \left(3 y\right)^{4-2} \left(\frac{x}{2}\right)^{2}=\frac{27}{2} x^{2} y^{2}$[/tex]
For [tex]k=3[/tex]
[tex]$\binom{4}{3} \left(3 y\right)^{4-3} \left(\frac{x}{2}\right)^{3}=\frac{3}{2} x^{3} y$[/tex]
For [tex]k=4[/tex]
[tex]$\binom{4}{4} \left(3 y\right)^{4-4} \left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{16}$[/tex]
You can perform the calculations, I will not type everything.
The answer is the sum of elements calculated.
Just organizing:
[tex]$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$[/tex]
