The n th term (Un) of a progression geometric is given by the formula
2(1.1)^n. Find the least value of n such that Un ≥ n
To find the value of n for which Un ≥ n, we need to solve the inequality 2(1.1)^n ≥ n.
Let's start by simplifying the inequality. We can divide both sides by n (assuming n is a positive number):
2(1.1)^n / n ≥ 1
Now, let's consider the left side of the inequality. The expression 2(1.1)^n / n can be simplified using exponential properties:
2 * (1.1^n / n)
Since we are trying to find the least value of n, let's check the values of n starting from 1 and going higher until we find a value that satisfies the inequality.
For n = 1:
2 * (1.1^1 / 1) = 2.2
Since 2.2 ≥ 1, n = 1 satisfies the inequality. However, we are looking for the least value of n, so we need to continue checking.
For n = 2:
2 * (1.1^2 / 2) = 2.42
Since 2.42 ≥ 2, n = 2 also satisfies the inequality. Again, we need to continue checking for smaller values of n.
For n = 3:
2 * (1.1^3 / 3) = 2.86
Since 2.86 ≥ 3, n = 3 satisfies the inequality as well.
For n = 4:
2 * (1.1^4 / 4) = 3.08
Since 3.08 ≥ 4, n = 4 satisfies the inequality.
As we can see, n = 1, 2, 3, and 4 all satisfy the inequality. However, we're looking for the least value of n. Therefore, the least value of n that satisfies the inequality is n = 1.