Use the binomial theorem to expand the binomial. $$\left( \frac { x } { 2 } + y \right) ^ { 4 }$$
A) $\frac { 1 } { 16 } x ^ { 4 } + \frac { 1 } { 8 } x ^ { 3 } y + \frac { 1 } { 4 } x ^ { 2 } y ^ { 2 } + \frac { 1 } { 2 } x y ^ { 3 } + y ^ { 4 }$
B) $\quad x ^ { 4 } + x ^ { 3 } y + x ^ { 2 } y ^ { 2 } + x y ^ { 3 } + y ^ { 4 }$
C) $\quad \frac { 1 } { 16 } x ^ { 4 } + \frac { 1 } { 2 } x ^ { 3 } y + \frac { 3 } { 2 } x ^ { 2 } y ^ { 2 } + 2 x y ^ { 3 } + y ^ { 4 }$
D) $\frac { 1 } { 8 } x ^ { 4 } + \frac { 1 } { 2 } x ^ { 3 } y + \frac { 3 } { 2 } x ^ { 2 } y ^ { 2 } + 2 x y ^ { 3 } + y ^ { 4 }$
E) $\frac { 1 } { 2 } x ^ { 4 } + y ^ { 4 }$

Question

Quizplus · Accepted Answer

We can expand the binomial using the binomial theorem formula:

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

In this case, $a=\frac{x}{2}$ and $b=y$, so we have:

$$\left(\frac{x}{2}+yight)^4 = \sum_{k=0}^{4} \binom{4}{k} \left(\frac{x}{2}ight)^{4-k} y^k$$

Let's simplify the binomial coefficients and powers of $\frac{x}{2}$:

$$\begin{aligned} \left(\frac{x}{2}+yight)^4 &= \binom{4}{0} \left(\frac{x}{2}ight)^4 y^0 + \binom{4}{1} \left(\frac{x}{2}ight)^3 y^1 + \binom{4}{2} \left(\frac{x}{2}ight)^2 y^2 + \binom{4}{3} \left(\frac{x}{2}ight)^1 y^3 + \binom{4}{4} \left(\frac{x}{2}ight)^0 y^4 \ &= \frac{1}{16} x^4 + \frac{1}{2} x^3 y + \frac{3}{4} x^2 y^2 + \frac{3}{2} x y^3 + y^4 \end{aligned}$$

Therefore, the correct choice is C.

Use the Binomial Theorem to Expand the Binomial (x2+y)4\left( \frac { x } { 2 } + y \right) ^ { 4 }(2x​+y)4

Use the Binomial Theorem to Expand the Binomial $\left( \frac { x } { 2 } + y \right) ^ { 4 }$