Sure, I can help with that. However, memoization is not typically used with tail recursion because memoization involves storing previous results to avoid duplicate computations. This is often done in the context of dynamic programming, not tail recursion.

However, I can show you how to write a tail-recursive version of the Fibonacci function and another version that uses memoization. Here's how you can implement both of these:

1. Tail recursive version

```c
int fib_tail(int n, int a, int b) {
    if (n == 0)
        return a;
    else if (n == 1)
        return b;
    else
        return fib_tail(n - 1, b, a + b);
}

int fib(int n) {
    return fib_tail(n, 0, 1);
}
```

In the above code, fib_tail is the actual tail-recursive function. It carries the last two Fibonacci numbers in a and b through each recursive call.

2. Memoized version

```c
#include <stdio.h>
#include <stdlib.h>

int* f;

int fib(int n) {
    if (n <= 1) {
        f[n] = n;
        return n;
    } else {
        if (f[n - 2] == -1)
            f[n - 2] = fib(n - 2);
        if (f[n - 1] == -1)
            f[n - 1] = fib(n - 1);
        return f[n - 2] + f[n - 1];
    }
}

int main() {
    int n = 10; // Change this to your desired Fibonacci term
    f = (int*) malloc((n + 1) * sizeof(int));
    for (int i = 0; i <= n; i++)
        f[i] = -1;
    printf("%d\n", fib(n));
    free(f);
    return 0;
}
```

This code uses an array f to store the Fibonacci numbers calculated so far. If a Fibonacci number is already calculated, it gets that number from the array instead of calculating it again. This is a basic implementation and does not include error checks for the memory allocation.
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