Question

Find the least value of n such that the sum 3+6+12+24+.... to n terms would fig exceed 1.5 million.​

Answer 1

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