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 January 14th, 2013, 04:00 AM #1 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 sum of the first 50 terms Determine the sum of the first 50 terms in the series 1+(1/2)+(1/2)^2 +(1/2)^3 +..... I actually already answered this question but would like to get confirmation on it! I used formula: Sn=(a1+an)/2 n After futher calculations: an=a1+(n-1) r , r=1/2, a50=? a50= 51/2 And then using main formula S50= 1325/2=662.5 Am i correct?
 January 14th, 2013, 05:06 AM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: sum of the first 50 terms The sum should be bounded above by 2, right?
 January 14th, 2013, 05:17 AM #3 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 Re: sum of the first 50 terms Yes, but i thought i already did that. S50= ((1+25.5)/ 2)* 50= 53/2 *25= 1325/2= 662.5
January 14th, 2013, 05:59 AM   #4
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Re: sum of the first 50 terms

Hello, Agata78!

Quote:
 $\text{Determine the sum of the first 50 terms: }\:S \;=\;1\,+\,\frac{1}{2}\,+\,\left(\frac{1}{2}\right )^2\,+\,\left(\frac{1}{2}\right)^3\,+\, \cdots$

$\text{W\!e have: }\:S_{50} \;=\;1\,+\,\frac{1}{2}\,+\,\left(\frac{1}{2}\right )^2\,+\,\left(\frac{1}{2}\right)^3\,+\, \cdots\,+\,\left(\frac{1}{2\right)^{49}$

$\text{This is a geometric series with: }\,a= 1,\;r = \frac{1}{2},\;n = 50$

$\text{Formula: }\:S_n \;=\;a\,\frac{1\,-\,r^n}{1-r}$

$\text{Hence: }\:S_{50} \;=\;1\,\cdot\,\frac{1-\left(\frac{1}{2}\right)^{50}}{1-\frac{1}{2}} \;=\;\frac{1\,-\,\frac{1}{2^{50}}}{\frac{1}{2}} \;=\;2\left(\frac{2^{50}\,-\,1}{2^{50}}\right)$

[color=beige]n . . . . .[/color]$S_{50} \;=\;\frac{2^{50}\,-\,1}{2^{49}} \;=\;2\,-\,\frac{1}{2^{49}}$

January 14th, 2013, 06:12 AM   #5
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Re: sum of the first 50 terms

Quote:
 Originally Posted by Agata78 Yes, but i thought i already did that. S50= ((1+25.5)/ 2)* 50= 53/2 *25= 1325/2= 662.5
662.5 > 2, so this clearly can't be the answer. soroban has the right idea, above.

 January 14th, 2013, 06:15 AM #6 Member   Joined: Jan 2013 Posts: 93 Thanks: 0 Re: sum of the first 50 terms I have completely different formula. But i think yours makes more sense then mine! Thank you

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