+4 Compute the value of \[\dfrac{1}{5} + \dfrac{3}{25} + \dfrac{7}{125} + \dfrac{15}{625} + \dfrac{31}{3125} + \cdots.\]
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.
+118680
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
what about the top?
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.
