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

Given: $2.25 \leq 2+2^{n+2} \leq 202$

We isolate the exponential term:

$0.25 \leq 2^{n+2} \leq 200$

Key insight: $0.25 = \tfrac{1}{4} = \tfrac{1}{2^2} = 2^{-2}$

So our inequality becomes:

$2^{-2} \leq 2^{n+2} \leq 200$

---

For the left inequality: $2^{-2} \leq 2^{n+2}$

Since the exponential function $2^x$ is increasing, we can compare exponents directly:

$-2 \leq n+2$

Therefore: $n \geq -4$

For the right inequality: $2^{n+2} \leq 200$

We need: $n+2 \leq \log_2(200)$

Since $2^7 = 128 < 200 < 256 = 2^8$, we have $7 < \log_2(200) < 8$

More precisely, $\log_2(200) \approx 7.64$

Since we need integers, $n+2 \leq 7$, so $n \leq 5$

Therefore: $-4 \leq n \leq 5$

This means $n \in \{-4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$

---

We calculate $3+3^{n+1}$ for each value of $n$:

$n$	$n+1$	$3^{n+1}$	$3+3^{n+1}$	Integer?
-4	-3	$\tfrac{1}{27}$	$3 + \tfrac{1}{27} = \tfrac{82}{27}$	No
-3	-2	$\tfrac{1}{9}$	$3 + \tfrac{1}{9} = \tfrac{28}{9}$	No
-2	-1	$\tfrac{1}{3}$	$3 + \tfrac{1}{3} = \tfrac{10}{3}$	No
-1	0	$1$	$3 + 1 = 4$	Yes
0	1	$3$	$3 + 3 = 6$	Yes
1	2	$9$	$3 + 9 = 12$	Yes
2	3	$27$	$3 + 27 = 30$	Yes
3	4	$81$	$3 + 81 = 84$	Yes
4	5	$243$	$3 + 243 = 246$	Yes
5	6	$729$	$3 + 729 = 732$	Yes

The integer values of $3+3^{n+1}$ are: $\{4, 6, 12, 30, 84, 246, 732\}$