Writing a series using Sigma notation How do I write $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}?$$
I have been struggling with these types of problems, so please, an explanation of how to get the result will be appreciated.

$$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
Since everything else has a denominator, let's try putting a denominator for $2$. Since anything divided by one itself, we get:
$$\frac{2}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
Everything else has an $x$. Let's give $2$ an $x$ too! $x^0 = 1$, and anything multiplied by $1$ is itself, so:
$$\frac{2x^0}{1}+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
All the other $x$s have an exponent. Let's give one to that $x$ next to the $3$. $x^1 = x$, so we get:
$$\frac{2x^0}{1}+ \frac{3x^1}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$
However, these denominators are just powers of two. Therefore, we have:
$$\frac{2x^0}{2^0}+ \frac{3x^1}{2^1} + \frac{4x^2}{2^2}+\frac{5x^3}{2^3}+\frac{6x^4}{2^4}$$
Now we see some patterns:


*

*$2, 3, 4, 5, 6$ (numerator of coefficient of $x$)

*$0, 1, 2, 3, 4$ (exponents of $x$)

*$1, 2, 4, 8, 16$ (denominator of coefficient of $x$. Powers of $2$, viz. $2^0, 2^1, 2^2, 2^3, 2^4$)


We'll make the sum range from $0$ to $4$, although in theory we could actually make it range over anything. 
$$\sum_{k=0}^4 $$
We need exponents of $x$, so we'll put those in there.
$$\sum_{k=0}^4 x^k $$
We need those powers of two in the denominator, so add that too:
$$\sum_{k=0}^4 \frac{x^k}{2^k} $$
However, there are still those numerators we haven't checked off our list yet. We can't have them ranging from $0$ to $4$, we need them from $2$ to $6$. The solution is to add two!
$$\sum_{k=0}^4 \frac{(k+2)x^k}{2^k} $$
