a direct sale company started out with two members in the first generation. each member has to recruit another two new members him or her. find the minimum number of generations such that the total number of members exceeds 1200
(hint: geometric series)
look at the sequence:
2
2+4=6
6+12 = 18
so this is a GP with a=2, r=3
2(3^n - 1)/(3-1) > 200
3^n > 201
n > 4.8
so, 5 generations
To find the minimum number of generations such that the total number of members exceeds 1200, we can use a geometric series.
Each member recruits two new members, so the number of members in each generation follows a geometric sequence with a common ratio of 2. The formula for the sum of a geometric series is:
Sn = a * (r^n - 1) / (r - 1)
Where:
- Sn is the sum of the geometric series
- a is the first term of the series
- r is the common ratio
- n is the number of terms (generations in this case)
Given that there are two members in the first generation (a = 2), we need to find the minimum value of n such that Sn > 1200.
Putting the given values into the formula:
1200 < 2 * (2^n - 1) / (2 - 1)
Now, we can simplify the inequality:
1200 < 2^n - 1
Add 1 to both sides:
1201 < 2^n
Take the logarithm base 2 of both sides:
log2(1201) < log2(2^n)
log2(1201) < n
Therefore, the minimum value of n that satisfies the inequality is ceil(log2(1201)).
Using a calculator, we find that ceil(log2(1201)) = 11.
So, the minimum number of generations required for the total number of members to exceed 1200 is 11.
To find the minimum number of generations in the direct sales company where the total number of members exceeds 1200, we can use a geometric series.
In the first generation, there are 2 members.
In the second generation, each of the 2 members in the first generation recruits 2 new members. So, there will be a total of 2 + (2 * 2) = 6 members in the second generation.
In the third generation, each of the 4 members in the second generation recruits 2 new members, resulting in a total of 6 + (4 * 2) = 14 members.
The pattern continues, with each generation having twice the number of members as the previous generation plus the previous generation's total.
So, in general, the number of members in each generation follows a geometric series with the first term (a) being 2 and a common ratio (r) of 2.
The formula for the sum of a geometric series is:
S = a * (1 - r^n) / (1 - r),
where S is the sum of the series, a is the first term, r is the common ratio, and n is the number of terms.
We need to find the value of n when the sum of the series (S) exceeds 1200.
Let's solve the equation:
1200 < a * (1 - r^n) / (1 - r).
Plugging in the given values, we have:
1200 < 2 * (1 - 2^n) / (1 - 2).
Simplifying further:
600 < (1 - 2^n).
Since the common ratio (r) is less than 1, the terms of the geometric series decrease as the generation increases. Therefore, to find the minimum number of generations, we need to find the highest possible value of n that satisfies the inequality.
Let's solve for n:
1 - 2^n > 600.
Subtracting 1 from both sides:
-2^n > 599.
Multiplying both sides by -1:
2^n < -599.
Since 2 raised to any power is always positive, there are no values of n that satisfy this inequality. Therefore, it is not possible for the total number of members to exceed 1200 in this scenario.