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I am in the process of learning to find general formulas for the summation of sequences. I've been given the general geometric sum

$$\sum_{i=m}^n r^i = \frac{r^m - r^{n+1}}{1 - r}$$

along with the basic summation properties, and am asked to find

$$\sum_{i=m}^n (-1)^{i^2}x^{2i}$$

Aside from rewriting the sum in different ways using the laws of exponents, I am stumped on finding the general equation for this sum and would enjoy any clue or solution.

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  • $\begingroup$ But you should use laws of exponents to rewrite, until it looks like $\sum (\text{something})^i$. I think $x^{2i}$ shouldn't be tough, but you'll have to think about $(-1)^{i^2}$... $\endgroup$ – pjs36 Jan 16 '17 at 4:06
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The parity of $i^2$ is the same as $i$, hence $$\sum_{i=m}^n (-1)^{i^2}x^{2i}=\sum_{i=m}^n (-1)^{i}x^{2i}$$ then use the general geometric sum we get \begin{align*} \sum_{i=m}^n (-1)^{i^2}x^{2i}&=\sum_{i=m}^n (-1)^{i}x^{2i}=\sum_{i=m}^n \left ( -x^{2} \right )^{i}\\&= \frac{\left ( -x^2 \right )^m - \left ( -x^2 \right )^{n+1}}{1 - \left ( -x^2 \right )}\\ &=\frac{\left ( -1 \right )^{m}x^{2m}+\left ( -1 \right )^{n}x^{2n+2}}{1+x^2} \end{align*}

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Hint: $i$ is even iff $i^2$ even so rewrite $(-1)^{i^2}$ as $(-1)^{i}$ then write $x^{2i}$ as $(x^2)^{i}$ and you can now rewrite everything as $(-(x^2))^i$, which you can use the given formula to evaluate.

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