How do you find the sum of the finite geometric sequence of #sum_(n=0)^50 10(2/3)^(n-1)#?

1 Answer
May 18, 2018

Answer:

#S=30[3/2+(2/3)^50]#

Explanation:

We know that,

"The sum of the first n-term of geometric sequence is :

#color(red)(S_n=(a(1-r^n))/(1-r)),where,a=# first term, #r=# common ratio "

We take,

#S=sum_(n=0)^50 10(2/3)^(n-1)=10sum_(n=0)^50(2/3)^(n-1)#

#=10sum_(n=0) (2/3)^(n-1)+10sum_(n=1)^50(2/3)^(n-1)#

=#10(2/3)^-1+10[(2/3)^0+(2/3)^1+(2/3)^2+...+(2/3)^49]#

#=10(3/2)+10[1+(2/3)^1+(2/3)^2+...+(2/3)^49]#

#:.S=10(3/2)+10*S'...to(I)#

Where, #S'=1+(2/3)^1+(2/3)^2+(2/3)^3+...+(2/3)^49#

Here,

#a=1, r=2/3 and n=1+49=50#

Using , #color(red)(S_n=(a(1-r^n))/(1-r)#

#:.S'=S_50=(1(1-(2/3)^50))/(1-2/3)#

#=>S'=(1-(2/3)^50)/((3-2)/3#

#=>S'=(1-(2/3)^50)/(1/3)#

#=>S'=3(1-(2/3)^50)#

Thus, from #(I)# we have

#S=10(3/2)+10[3(1-(2/3)^50)]#

#=30/2+30[1-(2/3)^50]#

#=30[1/2+1-(2/3)^50]#

#=30[3/2+(2/3)^50]#