# Math sequences

posted by Lala

Determine the smallest interger n such that 1+(4/5)+(4/5)^2+...+(4/5)^n>4.9

1. 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

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