I have this geometric series $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$to solve. So I extract the number two and get $2(\frac{1}{2}^0+ \frac{1}{2}^1+...+ \frac{1}{2}^7)$

I use the following formula $S_n= \frac{x^{n+1}-1}{x-1}$ so I plug in the values in this formula and get $S_n= 2\frac{\frac{1}{2}^{7+1}-1}{\frac{1}{2}-1}$ but the result is not correct.

What did I do wrong?


  • 1
    $\begingroup$ Are you sure that your final term is $2^{-7}$? Remember that you force-factored out 2... $\endgroup$ – baudolino Sep 27 '14 at 13:44
  • $\begingroup$ Your answer should be $2+\frac{\frac{1}{2}^{7+1}-1}{\frac{1}{2}-1}$ $\endgroup$ – Kim Jong Un Sep 27 '14 at 13:45
  • $\begingroup$ Last term would be $(1/2)^8$, inside the brackets (after taking 2 as common) $\endgroup$ – user141561 Sep 27 '14 at 13:45

$$S=2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$ Now if we subtract $2$ from both sides of the equation we get $$S-2=1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$ $$S-2=\frac{1}{2}\cdot \left(2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{64}\right)$$ Now notice that $2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{64}$ is $S-\frac1{128}$ so:

$$2S-4=S-\frac1{128} \rightarrow S=4-\frac1{128}=\frac{511}{128}$$


Hint: Its $$2 + (1+\frac12+ \frac14+ \cdots + \frac{1}{128})$$ not multiplied with $2$.

You can also think of it as follows: The first term is $a_1=2$ and the common ratio is $r=1/2$ and then you sum it using the formula where you last term is $a_9=1/128$.

Edit: If you do want to factor out a $2$, then you get $$2(1+\frac12 + \frac14 + \cdots + \frac{1}{256})= 2(\frac{1}{2^0} + \frac{1}{2^1} + \frac{1}{2^2} + \cdots + \frac{1}{2^8})$$

  • $\begingroup$ This is why it's wrong $\endgroup$ – ShakesBeer Sep 27 '14 at 14:38
  • $\begingroup$ @Shakespeare: the OP's method you mean? $\endgroup$ – Sheheryar Zaidi Sep 27 '14 at 14:40
  • $\begingroup$ Yes, exactly (well the OP's method was fine, he just made a slip), the answer is fine $\endgroup$ – ShakesBeer Sep 27 '14 at 14:42
  • $\begingroup$ Indeed, it's beneficial for the OP to know why their method is wrong rather than just get an answer for how to solve it. $\endgroup$ – Sheheryar Zaidi Sep 27 '14 at 14:44

You have


  • $\begingroup$ Someone didn't like this answer, why? $\endgroup$ – Timbuc Sep 27 '14 at 14:17
  • $\begingroup$ All you say seems to have been clear to the OP, he just made a mistake when extracting $\;2\;$ out of the original expression: he knows about geometric sequences, their sum and etc. No need to repeat all that but only, imo, to remark his mistake when taking out $\;2\;$ . $\endgroup$ – Timbuc Sep 27 '14 at 15:26

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