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I'm currently trying to rewrite a power series I have into summation notation.

The series is as follows: $$ 2x + 3x^{4} + 2x^{7} + 3x^{10} + 2x^{13} + ... $$

Obviously I'll have $x^{3n+1}$ in the summation, but I'm not sure on how to piece together the coefficient for each term.

I've worked with alternating coefficients before, typically when the coefficients can use the $(-1)^n$ trick in order to alternate between two specific integers, but I've never encountered a series when the coefficients differ by 1 each time. I feel like I'm overlooking something really simple in regards to solving this.

Thanks for any pointers or help.

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The coefficients are alternating between $2$ and $3$, so you can do something like

$$c_k = \frac{5+(-1)^k}{2}$$

Now you just have to express $k$ in terms of $n$, I think $k=3n+1$ might work.

We obviously only care whether $k$ is even or odd. So if $n$ is even, $k$ will be odd and vice versa. So we could also choose $k=n+1$.

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  • $\begingroup$ Awesome, just what I was looking for. I'll be sure to remember this division method to work with alternating series in the future. Thank you very much! $\endgroup$
    – Sierra117
    Jan 27 '16 at 21:17
  • $\begingroup$ It really is a simple trick, you can obviously construct sequences this way that alternate between any two values. $\endgroup$
    – flawr
    Jan 27 '16 at 21:21
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If we take the sum from $n=0$ on, then the exponent on the $n$th term is $3n+1$ as you said.

I'd just "construct" the coefficient. Start with a power of $-1$ to get the alternating behavior.

Since we start low with $n=0$, $(-1)^{n+1}$ gives us $-1, +1, -1, +1, ...$

If we multiply this by $\frac{1}{2}$ to decrease the difference from $2$ to $1$ (what we want) then we get $-\frac{1}{2}, +\frac{1}{2}, -\frac{1}{2}, +\frac{1}{2}, ...$

Now add $\frac{5}{2}$ and we've got it: $2, 3, 2, 3, ...$.

So,

$$S = \sum_{n=0}^{\infty} \left(\frac{5}{2} + \frac{1}{2}(-1)^{n+1}\right)x^{3n+1}.$$

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