Ask a New Question
Menu

how is this series 3+ 9/4 +27/16 +81/64... is converge to 12?

i represent the series as 3^(n+1)/4^n n starts from 0 to infinity, is this correct formula

2 answers

3*[1 + 3/4 + (3/4)^2 + ...(3/4)^n]
As n-> infnity, this becomes
3*[1/(1 - (3/4)] = 3*4 = 12
As n gets larger the sum goes to 0
Similar Questions
  1. Consider the infinite geometric series∑^(∞)_(n=1) −4(1/3)^n−1 . In this image, the lower limit of the summation notation
    1. answers icon 2 answers
  2. Consider the infinite geometric series∑ ∞ n = 1 − 4 ( 1 3 ) n − 1 . In this image, the lower limit of the summation
    1. answers icon 2 answers
  3. Consider the infinite geometric series∑ ∞ n = 1 − 4 ( 1 3 ) n − 1 . In this image, the lower limit of the summation
    1. answers icon 2 answers
  4. Consider the infinite geometric seriesn=1 infinity symbol -4(2/3)^n-1 a. Write the first four terms of the series. b. Does the
    1. answers icon 1 answer
more similar questions