To find the least value of n, we need to calculate the sum of the series until it exceeds 1.5 million. Let's start by writing the general term of the series:
a(n) = 3 x 2^(n-1)
We want to find the sum of the first n terms:
S(n) = 3 + 6 + 12 + ... + 3 x 2^(n-1)
Using the formula for the sum of a geometric series, we can write:
S(n) = 3(1 - 2^n)/(1 - 2)
Simplifying, we get:
S(n) = 3(2^n - 1)
Now we can set up an inequality to find the least value of n:
3(2^n - 1) > 1.5 x 10^6
2^n - 1 > 500000/3
2^n > 500001/3
n log 2 > log (500001/3)
n > log (500001/3) / log 2
n > 18.9316
Since n must be an integer, the least value of n that satisfies this inequality is 19. Therefore, the sum of the first 19 terms of the series exceeds 1.5 million
To find the least value of n, we need to calculate the sum of the series until it exceeds 1.5 million. Let's start by writing the general term of the series:
a(n) = 3 x 2^(n-1)
We want to find the sum of the first n terms:
S(n) = 3 + 6 + 12 + ... + 3 x 2^(n-1)
Using the formula for the sum of a geometric series, we can write:
S(n) = 3(1 - 2^n)/(1 - 2)
Simplifying, we get:
S(n) = 3(2^n - 1)
Now we can set up an inequality to find the least value of n:
3(2^n - 1) > 1.5 x 10^6
2^n - 1 > 500000/3
2^n > 500001/3
n log 2 > log (500001/3)
n > log (500001/3) / log 2
n > 18.9316
Since n must be an integer, the least value of n that satisfies this inequality is 19. Therefore, the sum of the first 19 terms of the series exceeds 1.5 million