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.
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
This question asks us to find the smallest integer value of 'n' such that the sum of a geometric series exceeds 4.9.
A geometric series is a sequence of numbers where each term is found by multiplying the previous term by a constant value called the common ratio. In this case, the common ratio is 4/5, which means each term is 4/5 times the previous term.
The general formula to find the sum of the first 'n' terms of a geometric series is given by:
Sn = a(1 - r^n) / (1 - r)
Where:
- Sn is the sum of the first 'n' terms
- a is the first term of the series
- r is the common ratio
In this case, the first term, 'a', is 1, and the common ratio, 'r', is 4/5. We want to find the smallest integer value of 'n' such that Sn exceeds 4.9.
To solve this problem, we can start by substituting the values into the formula and simplifying the expression:
1 + (4/5) + (4/5)^2 + ... + (4/5)^n > 4.9
Using the formula for the sum of a geometric series, we can write:
Sn = 1(1 - (4/5)^(n+1)) / (1 - 4/5)
Simplifying further:
5(1 - (4/5)^(n+1)) > 4.9
Now, let's solve this inequality:
1 - (4/5)^(n+1) > 4.9/5
- (4/5)^(n+1) > 3.9/5
Taking the logarithm of both sides to solve for 'n':
log((4/5)^(n+1)) < log(3.9/5)
(n+1) * log(4/5) < log(3.9/5)
n+1 > (log(3.9/5)) / log(4/5)
n > ((log(3.9/5)) / log(4/5)) - 1
Using a calculator to evaluate the right side of the inequality:
n > 15.11 - 1
n > 14.11
Since 'n' must be an integer, the smallest integer value that satisfies this inequality is n = 15.
Therefore, the smallest integer value of 'n' such that 1 + (4/5) + (4/5)^2 + ... + (4/5)^n > 4.9 is n = 15.