How do you determine the number of nth term in this geometric series #sum_{i=1}^n -4^(i-1)=-341#?

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Answer 1 · 2017-07-09T02:22:27

#sum_{i=1}^n -4^(i-1)=-341#

#=>sum_{i=1}^n 4^(i-1)=341#

#=>(4^0+4^1+4^2+--------4^(n-1)=341)#

#=>(4^0(4^n-1))/(4-1)=341#

#=>4^n=341xx3+1=1024=4^5#

#=>n=5#

Answer 2 · 2017-07-09T02:27:47

#n=5#

#sum_"i=1"^n -4^(i-1)# is a geometric sum.

The first term #(a_1) = -4^0 =-1#

The common ratio #(r) = (-4^1)/(-1) = 4#

The sum of the first n terms #(S_n)# is given by:

#S_n = (a_1(1-r^n))/(1-r)#

In this question we are told that #S_n = -341# and asked to solve for #n#.

#-341 = (-1(1-4^n))/(1-4)#

#(1-4^n)/3 = 341#

#1-4^n = -1023#

#4^n = 1024#

#2^(2n) = 2^10#

#2n =10 -> n=5#