3

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?

Thanks!!

S4M1R
  • 701

3 Answers3

6

$$S=2+1+ \frac{1}{2}+ \frac{1}{4}+...+ \frac{1}{128}$$ 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:

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

PunkZebra
  • 1,595
2

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 $$2(1+\frac12 + \frac14 + \cdots + \frac{1}{256})= 2(\frac{1}{2^0} + \frac{1}{2^1} + \frac{1}{2^2} + \cdots + \frac{1}{2^8})$$

Sheheryar Zaidi
  • 2,714
1

You have

$$2+\sum_{k=0}^7\frac1{2^k}=2+\frac{1-\frac1{2^8}}{1-\frac12}$$

Timbuc
  • 34,191
  • Someone didn't like this answer, why? – Timbuc Sep 27 '14 at 14:17
  • 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 $;2;$ . – Timbuc Sep 27 '14 at 15:26