Question

Use the bionomial theorem to write the binomial expansion

Answer 1

Answer:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

Step-by-step explanation:

$\left(\frac{1}{2}x+3y \right)^4=\left(\frac{x}{2}+3y \right)^4$

Binomial Expansion Formula:

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$, also $\binom{n}{k}=\frac{n!}{(n-k)!k!}$

We have to solve $\left(\frac{x}{2} + 3 y\right)^{4}=\sum_{k=0}^{4} \binom{4}{k} \left(3 y\right)^{4-k} \left(\frac{x}{2}\right)^k$

Now we should calculate for $k=0, k=1, k=2, k=3 \text{ and } k =4;$

First, for $k=0$

$\binom{4}{0} \left(3 y\right)^{4-0} \left(\frac{x}{2}\right)^{0}=\frac{4!}{(4-0)! 0!}\left(3 y\right)^{4} \left(\frac{x}{2}\right)^{0}=\frac{4!}{4!}(81y^4)\cdot 1 =81 y^{4}$

It is the same procedure for the other:

For $k=1$

$\binom{4}{1} \left(3 y\right)^{4-1} \left(\frac{x}{2}\right)^{1}=54 x y^{3}$

For $k=2$

$\binom{4}{2} \left(3 y\right)^{4-2} \left(\frac{x}{2}\right)^{2}=\frac{27}{2} x^{2} y^{2}$

For $k=3$

$\binom{4}{3} \left(3 y\right)^{4-3} \left(\frac{x}{2}\right)^{3}=\frac{3}{2} x^{3} y$

For $k=4$

$\binom{4}{4} \left(3 y\right)^{4-4} \left(\frac{x}{2}\right)^{4}=\frac{x^{4}}{16}$

You can perform the calculations, I will not type everything.

The answer is the sum of elements calculated.

Just organizing:

$\left(\frac{x}{2} + 3 y\right)^{4}=\frac{x^{4}}{16} + \frac{3}{2} x^{3} y + \frac{27}{2} x^{2} y^{2} + 54 x y^{3} + 81 y^{4}$

Answer 2

Answer:  $\bold{\dfrac{1}{16}x^4 + \dfrac{3}{2}x^3y + \dfrac{27}{2}x^2y^2 +54xy^3+81y^4}$

Step-by-step explanation:

                     Binomial Tree

n=0                         1

n=1                      1      1

n=2                  1    2     1

n=3               1    3    3     1

n=4            1    4    6    4    1

Using the Binomial Theorem

$\bigg(\dfrac{1}{2}x+3y\bigg)^4\\\\\\=1\bigg(\dfrac{1}{2}x\bigg)^4(3y)^0\quad \rightarrow \quad \dfrac{1}{16}x^4\\\\+4\bigg(\dfrac{1}{2}x\bigg)^3(3y)^1\quad \rightarrow \quad \dfrac{3}{2}x^3y\\\\+6\bigg(\dfrac{1}{2}x\bigg)^2(3y)^2\quad \rightarrow \quad \dfrac{27}{2}x^2y^2\\\\+4\bigg(\dfrac{1}{2}x\bigg)^1(3y)^3\quad \rightarrow \quad 54xy^3\\\\+1\bigg(\dfrac{1}{2}x\bigg)^0(3y)^4\quad \rightarrow \quad 81y^4$

______________________

$= \dfrac{1}{16}x^4 + \dfrac{3}{2}x^3y + \dfrac{27}{2}x^2y^2 +54xy^3+81y^4$