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I've got this: $$\sum_{k=1}^{n}\frac{1}{2^{2k}}$$Wolfram Alpha already simplified it for me as $\dfrac{2^{2n}-1}{3\cdot2^{2n}}$, but he hasn't told me why.

k5f
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    Would it help to note $\frac{1}{2^{2k}} = \left(\frac{1}{4}\right)^k$? – Jonathan Y. Dec 02 '13 at 19:59
  • I see that too, but I don't know what's $\sum_{k=1}^{n}x^k$ equal to. – k5f Dec 02 '13 at 20:01
  • @MathNoob, the trick is to multiply the sum by $(1-x)$ (if you do it once yourself and arrive by the familiar formula, I'd bet you'll remember it from thereon out.) – Jonathan Y. Dec 02 '13 at 20:06

2 Answers2

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For any geometric series

$$\sum_{k=1}^n c^k=c\frac{c^n-1}{c-1}$$

so in your case, as Jonathan's comment proposes

$$\sum_{k=1}^n\frac1{2^{2k}}=\sum_{k=1}^n\left(\frac14\right)^k=\frac14\frac{\left(\frac14\right)^n-1}{\frac14-1}=\frac{\frac{1-4^n}{4^n}}{1-4}=\frac{2^{2n}-1}{3\cdot2^{2n}}$$

DonAntonio
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$$\sum_{k=1}^{n}\frac{1}{2^{2k}}=\sum_{k=1}^{n}\left(\frac{1}{4}\right)^k=\frac{1}{4}\frac{1-(1/4)^{n}}{1-1/4}=\frac{1}{4}\frac{1-(1/4)^{n}}{3/4}$$ $$=\frac{1-(1/4)^{n}}{3}=\frac{4^{n}-1}{3\cdot4^{n}}=\frac{(2^2)^{n}-1}{3\cdot(2^2)^{n}}=\frac{2^{2n}-1}{3\cdot2^{2n}}$$

Adi Dani
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