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.


5 Answers 5


$$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}$$

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!

$$\sum_{k=0}^4 \frac{(k+2)x^k}{2^k} $$

  • $\begingroup$ I'd have gone on to write $\dfrac{2x^0}{1}+ \dfrac{3x^1}{2} + \dfrac{4x^2}{4}+\dfrac{5x^3}{8}+\dfrac{6x^4}{16}$ in the form $\dfrac{2x^0}{2^0}+ \dfrac{3x^1}{2^2} + \dfrac{4x^2}{2^2}+\dfrac{5x^3}{2^3}+\dfrac{6x^4}{2^4}$. $\endgroup$ Commented Apr 17, 2013 at 23:34
  • $\begingroup$ @MichaelHardy, good idea, I added it. $\endgroup$ Commented Apr 18, 2013 at 0:05

Note that the power of $x$ increase in steps of $1$ starting from $x^0$. Hence the $(n+1)^{th}$ term has $x$ raised to the power $n$. The denominators are powers of $2$ starting from $2^0$ and the numbers in the numerator also increase in steps of $1$ starting from $2$. $$\sum_{n=0}^4 \dfrac{(n+2)x^n}{2^n}$$


It's much like finding a pattern, and a way to express the pattern. Starting point: $2 = \dfrac 21$. Numerator coeffienct of the $x$-term, increases by one per term , starting at $2$ (add 1 to starting numerator); the exponent for $x$ increases by one (starting, say, at $x^0,...x^1 = x,...x^2...$ (Add one to the x-exponent starting at $0$. Also, the denominator represents powers of $2$, starting at $2^0 = 1$...Increase (add) by one the exponent of the term $2$ in the denominator $2^0, ... 2^1 = 2, ...2^{1+1} = 2^2 = 4, ...$

Try writing out the first few terms of the following and see if it matches, starting index $0$, and on...:

$$\sum_{k = 0}^\infty \frac{(k+2)x^k}{2^k}$$

If things don't work out (here they happen to work just fine), there is often a little trial-and-error involved. For example, we can "tweak" if we want to represent the series with a starting index $1$, by "subtracting $1$ from each of the expressions given in terms of $k$.

NOTE: The above will give you an infinite series. To get the first five terms, we start at $k = 0$, and need to sum through (inclusive) $k = 4$. But you can obtain any positive number $n$ of terms by summing from $k = 0$ to $k = n-1)$

  • $\begingroup$ The best way to get a "knack" for "cracking" a sequence is through practice. Try to approach sequences of numbers like you'd approach a puzzle: what fits? what doesn't fit? What patterns do I see...start writing down what you do see, and try creating a "model" term (a "work in progress," so to speak)...Figure out if starting with $0$ or $1$ for one of the varying terms works best. If a term appears constant through the sequence, represent it as a constant, etc. $\endgroup$
    – amWhy
    Commented Apr 17, 2013 at 23:47
  • $\begingroup$ Such great advice! +1 $\endgroup$
    – Amzoti
    Commented Apr 18, 2013 at 0:24
  • $\begingroup$ Such a great answer +1 $\endgroup$
    – Mikasa
    Commented Aug 21, 2013 at 6:14

Notice that $2=\frac {2x^0}1$, so you want $$2+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\frac {2x^0}1+ \frac{3x}{2} + \frac{4x^2}{4}+\frac{5x^3}{8}+\frac{6x^4}{16}=\sum_{i=1}^5\frac {(i+1)x^{i-1}}{2^{i-1}}$$


Start off by looking for any patterns. I would start off by looking at how the powers on $x$ change. Starting at $0$ ($x^0=1$) it looks like the power goes up by one at each "step." So we can start off with the "skeleton" sum $$ \sum_{i=0}^4 x^i.$$ Similarly observe the coefficient patters, it looks like the numerator starts at $2$ and goes up by one at each step and the denominator are powers of $2$ starting at the $2^0$. So putting that all together we get $$\sum_{i=0}^4 \frac{(i+2)x^i}{2^i}$$


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