For a GP
sum(n) = a(1 - r^n)/(1-r)
so for yours ....
sum(10) = -1( 1 - (1/3)^10)/(1 - 1/3)
= -(1 - 1/59049)/(2/3)
= -(59048/59049)(3/2) = ....
the following problem refers to a geometric sequence:
Find S_10 for -1, -1/3, -1/9
r=-1/3/-1=1/3
a_1=-1
a_2=-1/3
a_n=a_1*r^n-1
a_n=-1*1/3^n-1
s_n=
Help me.
1 answer