How do you find the sum of the finite geometric sequence of #Sigma 2(4/3)^n# from n=0 to 15?

Question

Evelyn Agard · Accepted Answer

#sum_(n=0)^15 = 2(4/3)^n# #2[(4/3)^o + (4/3)^1 + (4/3)^2...(4/3)^15]# Sum of any geometric finite geometric series is, #= (a(r^n-1))/(r-1)# if #r!=1# where, 
#a =# first term of the series
#r =# common ratio of the series substituting the values,   #2[(1((4/3)^16-1))/(4/3-1)]# #2[((4^16-3^16)/3^16)/(1/3)]# #2[(4^15-3^15)/3^15]# Symplify and get the answer.

William Abdalla · Answer

#sum_(n=0)^15 2(4/3)^n approx 592.647311# Sum #=sum_(n=0)^15 2(4/3)^n = 2sum_(n=0)^15 (4/3)^n# Now, #sum_(n=0)^15 (4/3)^n# is the sum of geometric progression, with 1st term #a=1# and common ratio #r=4/3# The sum of a finite GP #= (a(1-r^n))/(1-r)# where #n# is the number of terms. In our example, #a=1, r=4/3, n=16# #:. sum_(n=0)^15 (4/3)^n = (1*(1-(4/3)^16))/(1-4/3)# #approx -98.77455184xx -3# #approx 296.3236555# Sum #= 2 * sum_(n=0)^15 (4/3)^n # #approx 2xx 296.3236555# #approx 592.647311#

Emily Ammerman · Answer

undefined You can find the sum of the finite geometric sequence using the formula for the sum of a geometric series:

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

Where:
- $ S_n $ is the sum of the first $ n $ terms,
- $ a $ is the first term of the sequence,
- $ r $ is the common ratio,
- $ n $ is the number of terms.

For the given sequence $ \Sigma 2\left(\frac{4}{3}\right)^n $ from $ n = 0 $ to $ 15 $:
- $ a = 2 $ (the first term)
- $ r = \frac{4}{3} $ (the common ratio)
- $ n = 15 $ (number of terms)

Substitute these values into the formula and calculate $ S_{15} $ to find the sum of the sequence.