Question

For all possible integers n satisfying $2.25 ≤ 2 + 2^{n + 2} ≤ 202$, the number of integer values of $3 + 3^{n + 1}$ is

Answer 1

The inequality $2.25 ≤ 2 + 2^{n + 2} ≤ 202$ can be simplified to ensure that $2 + 2^{n + 2}$ is less than or equal to 202.

If we take 27 = 128, this satisfies the inequality, but 28 does not.

So, the maximum value for n is 5.

Additionally, when subtracting 2 from both sides of the inequality, we get $0.25 ≤ 2^{n + 2}$, which can be further simplified to $2 - 2 ≤ 2^{n + 2}$. This implies that $n ≥ -4.$

Therefore, the integral values of n can range from -4 to 5.

For the expression $3 + 3^{n + 1}$ to have integral values, n must be greater than or equal to -1 and go up to 5. Hence, the possible values of n are -1, 0, 1, 2, 3, 4, and 5.

So, there are 7 possible values for n.