Is there more than just one term?

Question

Find the fourth term of $(2+y)^{\frac{1}{2}}$

Wouldn't it just be $\sqrt{(2+y)}$? So isn't it just one term?

Answer 1

$$(a+b)^n=a^nb^0+\frac{n}1a^{n-1}b^1+\frac{n(n-1)}{1\times2}a^{n-2}b^2+\frac{n(n-1)(n-2)}{1\times2\times3}a^{n-3}b^3+\dots$$

Plugging in values and taking the fourth term:

$$\begin{align} 4th\text{ term of }(2+y)^{1/2} & =\frac{1/2(1/2-1)(1/2-2)}{1\times2\times3}2^{1/2-3}y^3 \\ & =\frac{\sqrt2}2y^3 \end{align}$$

Answer 2

Hint: $$(1+x)^{\frac{1}{2} }=\sum _{n=0}^{\infty }{\frac{1}{2} \choose n}x^{n}\quad {\text{ for all }}|x|<1$$ $${\alpha \choose n}=\prod _{k=1}^{n}{\frac {\alpha -k+1}{k}}={\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}.$$

$$(2+y)^{\frac{1}{2}}=\sqrt{2}(1+\frac{y}{2})^{\frac{1}{2}}$$