+0  
 
0
95
4
avatar+516 

Find the sum of \(1 - \frac{1}{2} + \frac{1}{5} - \frac{1}{10} + \frac{1}{25} - \frac{1}{50} + \frac{1}{125} - \frac{1}{250} + \cdots + \frac{1}{5^{10}} - \frac{1}{2 \cdot 5^{10}}\).

The answer should be in the form of a decimal rounded to the nearest HUNDREDTH.

 

I think this problem is looking for a few equations related to the sequence, but not sure which ones. Thanks :/

 Feb 27, 2022
 #1
avatar
0

I used a calculator, and got 0.65.

 Feb 27, 2022
 #2
avatar+117224 
+2

 

 

\(1 - \frac{1}{2} + \frac{1}{5} - \frac{1}{10} + \frac{1}{25} - \frac{1}{50} + \frac{1}{125} - \frac{1}{250} + \cdots + \frac{1}{5^{10}} - \frac{1}{2 \cdot 5^{10}}\\ =1 + \frac{1}{5} + \frac{1}{25} +\frac{1}{125} + \cdots + \frac{1}{5^{10}} \\ - \frac{1}{2} - \frac{1}{10} - \frac{1}{50} - \frac{1}{250} + \cdots - \frac{1}{2 \cdot 5^{10}}\)

 

the first line is the sum of a GP    a=1,          r=1/5       11 terms

the second is the sum of a GP     a=-1/2,      r=1/5      11 terms        (that dot is a times sign not a decimal point) 

 

Work them out and add them together.      

 Feb 27, 2022
 #3
avatar+516 
0

Thanks for the initial step Melody.

So for the first line I got 1.2499999744.

Then subtracting from the first line with the second line I got very close guess. Instead of subtracting everything else. I got to 0.6250015744, and I knew rounding to the nearest hundredth would get me 0.63 or 0.62. 

 

With your strategy, I got 0.62 as the nearest hundredth.

proyaop  Feb 28, 2022
 #4
avatar+117224 
+1

idk, I never finished it.

 

\(S_n=\frac{a(1-r^n)}{1-r}\\ S_{11}=\frac{1(1-0.2^{11})}{0.8}+\frac{-0.5(1-0.2^{11})}{0.8}\\ S_{11}=\frac{0.5(1-0.2^{11})}{0.8}\\ S_{11}=0.624999...\\ S_{11}\approx 0.62 \qquad \text{to 2 dec places}\)

Melody  Feb 28, 2022

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