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]#
