Finding General Formulas for Series: 1.01, 2.0301, 3.060401, 4.10100501...

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The discussion focuses on finding a general formula for the series defined by the partial sums S_n = sum of (1.01)^n from n=1 to infinity. Participants explore the pattern in the partial sums, noting that S1=1.01, S2=2.0301, S3=3.060401, and S4=4.10100501. Suggestions include comparing successive terms and utilizing properties of geometric series to derive a formula. The conclusion highlights that understanding the ratios of terms can lead to a successful generalization for S_n. Ultimately, the problem is resolved using the geometric series formula.
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The question is about the series:
sum of (1.01)^n from n=1 to infinity

It asks me to investigate the partial sums for that series and then find an expression for Sn. The partial sum part goes like:
S1=1.01
S2=1.01+(1.01)^2=2.0301
S3=1.01+(1.01)^2+(1.01)^3=3.060401
S4=1.01+(1.01)^2+(1.01)^3+(1.01)^4=4.10100501
etc

I'm stuck on finding the general formula (Sn) for the sequence
1.01, 2.0301, 3.060401, 4.10100501...
My friends and I have only gotten as far as guessing that it might be n+(1/something^something) but I'm wonderfing if there is a better way to find a formula for a sequence of numbers rather than guessing?
 
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Look up on geometric series.

Or compare S_n against 1.01S_n.
 
The answer should be apparent from the information you write at the start. In general, if you weren't already given what the sum is, you might was to consider looking at the following to see if provides any hints.

S(4) - S(3)
S(3) - S(2) etc.

See if any pattern arises which allows you to identify how successive terms are added.

After reading about geometric series you may find that forming

S(4) - S(3)/(S(3) - S(2))
S(3) - S(2)/(S(2) - S(1)) etc

Provides some help.
 
Last edited:
\sum_{n=1}^{\infty} A

A is a general expression of your pattern in terms of "n"

You need to find how to generalize the pattern for finding an nth term.
Look at the ratios of each term... you should have formulas for calculating a summation with the given ratio.
 
Thanks for that guys. I worked it out using the geometric series formula :)
 
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