How do you find the sum of the finite geometric sequence of #Sigma 5(3/5)^n# from n=0 to 40?

Question

William Abdalla · Accepted Answer

12.5 (1.d.p)

First calculate the first three terms: #5(3/5)^0= color(blue)(5) , 5(3/5)^1=color(blue)(3) , 5(3/5)^2=color(blue)(9/5)# Find the common ratio: #3/5 = (9/5)/3=3/5# The sum of a geometric sequence is: #a((1-r^n)/(1-r))# Where a is the first term, n is the nth term and r is the common ratio. So: #5((1-(3/5)^40)/(1-(3/5)))=12.4999999899743791#(16 .d.p.)

Christopher Agresta · Answer

undefined To find the sum of the finite geometric sequence $ \Sigma 5\left(\frac{3}{5}\right)^n $ from $ n = 0 $ to $ n = 40 $, you can use the formula for the sum of a finite geometric series:

$$ S_n = \frac{a(1 - r^{n+1})}{1 - r} $$

where:
$ a $ is the first term of the sequence,
$ r $ is the common ratio, and
$ n $ is the number of terms.

In this sequence, $ a = 5 $ (the first term) and $ r = \frac{3}{5} $ (the common ratio).

Plugging these values into the formula and substituting $ n = 40 $, you get:

$$ S_{40} = \frac{5(1 - \left(\frac{3}{5}\right)^{40+1})}{1 - \frac{3}{5}} $$

$$ S_{40} = \frac{5(1 - \left(\frac{3}{5}\right)^{41})}{\frac{2}{5}} $$

$$ S_{40} = 25\left(1 - \left(\frac{3}{5}\right)^{41}\right) $$

This will give you the sum of the finite geometric sequence.