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# Help desperately now!!!!! PLS whoever helps is amazing

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Compute the value of $\dfrac{1}{5} + \dfrac{3}{25} + \dfrac{7}{125} + \dfrac{15}{625} + \dfrac{31}{3125} + \cdots.$

Mar 11, 2023

#1
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The given series is a geometric series with first term a and common ratio r.

Here, a = 1/5 and r = 3/5.

Using the formula for the sum of an infinite geometric series with |r| < 1, we have:

Sum = a / (1 - r)

Sum = (1/5) / (1 - 3/5)

Sum = (1/5) / (2/5)

Sum = 1/2

Therefore, the sum of the given series is 1/2.

Mar 11, 2023
#2
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∑[5^-n * (-1 + 2^n), n, 1, ∞ ]==5 / 12 ==0.4166667

Mar 11, 2023
#3
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Hi Aiden

$$\dfrac{1}{5} + \dfrac{3}{25} + \dfrac{7}{125} + \dfrac{15}{625} + \dfrac{31}{3125} + \cdots.$$

I can see that the bottom is  5^n

1,3,7,15,31    the didifferences between these is  2,4,8,16    each being a increasing power of 2

so the top is     2^n - 1

so we have

$$\displaystyle \sum_{n=1}^{\infty}\;\;\frac{2^n-1}{5^n}\\~\\ =\displaystyle \sum_{n=1}^{\infty}\;\;\left(\frac{2}{5}\right)^n\;\;-\;\; \displaystyle \sum_{n=1}^{\infty}\;\;\left(\frac{1}{5}\right)^n\\$$

You should be able to take it from there.

Mar 12, 2023