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# HARD series problem help

+1
391
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+1622

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
-2

I used a calculator, and got 0.65.

Feb 27, 2022
#2
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$$1 - \frac{1}{2} + \frac{1}{5} - \frac{1}{10} + \frac{1}{25} - \frac{1}{50} + \frac{1}{125} - \frac{1}{250} + \cdots + \frac{1}{5^{10}} - \frac{1}{2 \cdot 5^{10}}\\ =1 + \frac{1}{5} + \frac{1}{25} +\frac{1}{125} + \cdots + \frac{1}{5^{10}} \\ - \frac{1}{2} - \frac{1}{10} - \frac{1}{50} - \frac{1}{250} + \cdots - \frac{1}{2 \cdot 5^{10}}$$

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
+1622
+2

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
+118617
+3

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