Consider the following C function.int fun(int n) { int x=1, k; if (n==1) return x; for (k=1; k<n; ++k) x = x + fun(k) * fun(n-k); return x; }The return value of fun(5) is _______.

We can compute the value returned by fun(n) recursively for increasing values of $n$:

1. For $n = 1$: The base condition if (n == 1) triggers, returning:

   $$\text{fun}(1) = 1$$

2. For $n = 2$: Initially $x = 1$.

   • For $k = 1$: $x = 1 + \text{fun}(1) \times \text{fun}(1) = 1 + (1 \times 1) = 2$

   $$\text{fun}(2) = 2$$

3. For $n = 3$: Initially $x = 1$.

   • For $k = 1$: $x = 1 + \text{fun}(1) \times \text{fun}(2) = 1 + (1 \times 2) = 3$

   • For $k = 2$: $x = 3 + \text{fun}(2) \times \text{fun}(1) = 3 + (2 \times 1) = 5$

   $$\text{fun}(3) = 5$$

4. For $n = 4$: Initially $x = 1$.

   • For $k = 1$: $x = 1 + \text{fun}(1) \times \text{fun}(3) = 1 + (1 \times 5) = 6$

   • For $k = 2$: $x = 6 + \text{fun}(2) \times \text{fun}(2) = 6 + (2 \times 2) = 10$

   • For $k = 3$: $x = 10 + \text{fun}(3) \times \text{fun}(1) = 10 + (5 \times 1) = 15$

   $$\text{fun}(4) = 15$$

5. For $n = 5$: Initially $x = 1$.

   • For $k = 1$: $x = 1 + \text{fun}(1) \times \text{fun}(4) = 1 + (1 \times 15) = 16$

   • For $k = 2$: $x = 16 + \text{fun}(2) \times \text{fun}(3) = 16 + (2 \times 5) = 26$

   • For $k = 3$: $x = 26 + \text{fun}(3) \times \text{fun}(2) = 26 + (5 \times 2) = 36$

   • For $k = 4$: $x = 36 + \text{fun}(4) \times \text{fun}(1) = 36 + (15 \times 1) = 51$

   $$\text{fun}(5) = 51$$

Thus, the return value of fun(5) is 51.