Asked by cecile
find the sum of the infinite series (2/9)^2-(2/9)^3+(2/9)^4-(2/9)^5+ ...
Answers
Answered by
Damon
(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)^2-1(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 (1-r^n)/(1-r)
n is infinite and |r| < 1
so Sn =g/(1-r)
Sn = 1/(1-2/9) = 1.286
so we want
1.286 -[1-2/9]
.508
This is like a geometric series with the first terms missing
1 -1(2/9)^1 +1(2/9)^2-1(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 (1-r^n)/(1-r)
n is infinite and |r| < 1
so Sn =g/(1-r)
Sn = 1/(1-2/9) = 1.286
so we want
1.286 -[1-2/9]
.508
Answered by
Damon
Whoa - typo
Sn = 1/(1+2/9) because r = -2/9
Sn = .818
so we want
.818 -[1-2/9]
.040
Sn = 1/(1+2/9) because r = -2/9
Sn = .818
so we want
.818 -[1-2/9]
.040
Answered by
ashlee and blake
i need help
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