Asked by Lala
Determine the smallest interger n such that 1+(4/5)+(4/5)^2+...+(4/5)^n>4.9
Please explain ti me about thisquestion. I can'tunderstand it.
Please explain ti me about thisquestion. I can'tunderstand it.
Answers
Answered by
Reiny
you have a geometric series with
a = 1 and r = 4/5
sum(n) = a(1 - r^n)/(1-r) > 4.9
1(1 - (4/5)^n)/1/5 > 4.9
1 - (4/5)^n > .98
(4/5)^n < .02
suppose we let .8^n = .02
log .8^n = log .02
n = log.02/log.8 = 17.53
testing:
if n = 17 , sum(17) = (1 - .8^17)/.2 = 4.8874 < 4.9
if n = 18 , sum(18) = (1-08^18)/.2 = 4.9099 > 4.9
So the smallest value of n is 18
a = 1 and r = 4/5
sum(n) = a(1 - r^n)/(1-r) > 4.9
1(1 - (4/5)^n)/1/5 > 4.9
1 - (4/5)^n > .98
(4/5)^n < .02
suppose we let .8^n = .02
log .8^n = log .02
n = log.02/log.8 = 17.53
testing:
if n = 17 , sum(17) = (1 - .8^17)/.2 = 4.8874 < 4.9
if n = 18 , sum(18) = (1-08^18)/.2 = 4.9099 > 4.9
So the smallest value of n is 18
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