Been working out series for the past day and have come across this question. I would kindly ask how I ought to tackle it, I've done Binomial and Maclauren's however this seems to require another method.
Given that x is so small that $x^3$ and higher powers of $x$ may be neglected, write down a quadratic approximation of each of the following: $$1. \frac{1+3x}{2+x} 2. \frac{2+5x}{(1-3x)^2}$$
Those happen to be the two questions to work out. I kindly ask to not give me the answer immediately but simply steer me to the right direction.
My Attempt(s)
Started off by rewriting everything$$\frac{1+3x}{2+x} = \frac{1}{2+x} + \frac{1}{2+x}3x$$
Then proceeded to factor 2 out to bring the it to the form $(1+y)$ $$\frac{1}{2(1+\frac{x}{2})} + \frac{1}{2(1+\frac{x}{2})}3x$$
For convenience sake (personal thing I guess) I rewrote it as:$$2^{-1}(1+\frac{x}{2})^{-1} + 2^{-1}(1+\frac{x}{2})^{-1}3x$$
And therefore expanded everything like so:$$\frac{1}{2}\left [ 1+(-1)(\frac{x}{2}) + \frac{(-1)(-2)(\frac{x}{2})^2}{2!} + ...\right ] + \frac{3x}{2}\left [1+(-1)(\frac{x}{2}) + \frac{(-1)(-2)(\frac{x}{2})^2}{2!} + ...\right ]$$
Simplified everything down to:$$\frac{1}{2}\left [ 1- \frac{x}{2} - \frac{x^2}{4} + ...\right ] + \frac{3x}{2}\left [ 1- \frac{x}{2} - \frac{x^2}{4} + ...\right ]$$
Disregarded any terms that where degree 3 or higher (there was only one): $$\frac{1}{2} - \frac{x}{4} - \frac{x^2}{8} + \frac{3x}{2} - \frac{3x^2}{4}$$
And thus finally arrived with: $$\frac{1}{2} - \frac{5x}{4} - \frac{5x^2}{8}$$
For the first question, note that $\frac{1+3x}{2+x} = \frac{1}{2+x} + 3x * \frac{1}{2+x}$. Now, try to put $\frac{1}{2+x}$ into the form $\frac{1}{1-y}$, and use the fact that, for $|y|<1$, we have $\frac{1}{1-y} = \sum_{n=0}^{\infty} y^n$.
The second question can be answered similarly. You just have to determine the series for $\frac{1}{(1-y)^2}$ using the series $\frac{1}{1-y} = \sum_{n=0}^{\infty} y^n$.
Let me know if you need more help.
Full Answer: We have $$\frac{1+3x}{2+x} = \frac{1+3x}{2} \frac{1}{1-(-\frac{x}{2})}.$$ The term on the right is equal to $$\frac{1}{1-(-\frac{x}{2})} = \sum_{n = 0}^{\infty} \left(-\frac{x}{2} \right)^n = 1 - \frac{x}{2} + \frac{x^2}{4} - \dots$$ Therefore, $$\frac{1+3x}{2} \frac{1}{1-(-\frac{x}{2})} = \frac{1+3x}{2} \left( 1 - \frac{x}{2} + \frac{x^2}{4} - \dots \right) \approx \frac{(1+3x)(4-2x+x^2)}{8} = \frac{4-2x+x^2+12x-6x^2+3x^3}{8} \approx \frac{-5x^2+10x+4}{8}.$$
This is equivalent to your answer except for the sign on $-\frac{5x}{4}$.