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

2 Answers
Jan 22, 2018

#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.

Jan 22, 2018

Answer:

#sum_(n=0)^15 2(4/3)^n approx 592.647311#

Explanation:

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#