maths
posted by cecile on .
find the sum of the infinite series (2/9)^2(2/9)^3+(2/9)^4(2/9)^5+ ...

(2/9)^2(2/9)^3+(2/9)^4(2/9)^5+
This is like a geometric series with the first terms missing
1 1(2/9)^1 +1(2/9)^21(2/9)^3+(2/9)^4(2/9)^5+
so if we find the sum of that series and subtract [1 1(2/9)^1], we have it
so
g = 1
r = 2/9
in Sn = g (1r^n)/(1r)
n is infinite and r < 1
so Sn =g/(1r)
Sn = 1/(12/9) = 1.286
so we want
1.286 [12/9]
.508 
i need help

Whoa  typo
Sn = 1/(1+2/9) because r = 2/9
Sn = .818
so we want
.818 [12/9]
.040