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I am not sure how to find a formula for the following series.

$\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\frac{31}{32}$

I am thinking it might be $\frac{2N-1}{N^2}$ for the numerator but this does hold on n=3

another series I am trying to find is

$2,0,2,0,2$

I kow $-1^n=-1$

But I a not sure what this could be

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    $\begingroup$ HINT: For the 1st series note that the numerator is always 1 less than the denominator and the denominator is a power of 2. $\endgroup$ – Mufasa Sep 27 '13 at 21:53
  • $\begingroup$ I see so it is $2^{n}-1/2^n$ $\endgroup$ – Fernando Martinez Sep 27 '13 at 21:55
  • $\begingroup$ Please be careful with parentheses. $2N-1/N^2=2N-\frac 1{N^2}\neq \frac {2N-1}{N^2}$. Clearly you mean the latter, but often it is not so clear. In your recent comment, the $-1$ should not be in the exponent. $\endgroup$ – Ross Millikan Sep 27 '13 at 21:56
  • $\begingroup$ Is there any way to know what pattern it is using a formula because I am bad at seeing these patterns..... $\endgroup$ – Fernando Martinez Sep 27 '13 at 21:58
  • $\begingroup$ the another series $1 - ?? = 0 \text{ or 2}$ ?? $\endgroup$ – Santosh Linkha Sep 27 '13 at 21:58
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If you meant for the nth term in the first case to be $\dfrac {2n-1}{2^n}$, your close, but let's try $$\dfrac{2^n - 1}{2^n}\quad ?$$

For the second, how about $$(1 - (-1)^n)\quad ?$$

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  • $\begingroup$ Yes that works perfectly $\endgroup$ – Fernando Martinez Sep 27 '13 at 22:06
  • $\begingroup$ thanks hmm I guess it takes practice to be able to see a formula from a series... $\endgroup$ – Fernando Martinez Sep 27 '13 at 22:10
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    $\begingroup$ Your welcome, Fernando. Yes, indeed, practice with all sorts of series helps you develop a lot more intuition for "pattern finding". $\endgroup$ – Namaste Sep 27 '13 at 22:13
  • $\begingroup$ Also, just to be clear: recall that $(-1)^n$ is $-1$ for odd $n$, but $1$ for even $n$. This makes sense because every even $n$ can be expressed as $2k$, so $(-1)^n = (-1)^{2k} = \Big((-1)^2\Big)^k = 1^k = 1$ where $k \in \mathbb N, k\geq 0, 1$, depending on indexing. A similar argument can be made for odd $n$. $\endgroup$ – Namaste Sep 27 '13 at 22:22
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HINT: The numerator is increasing by $2^n$ and the denominator by $2^n$.

But you have a number less on the top...so....$2^n-1$ could be the numerator perhaps? $2^n$ the denominator?

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