Consider the following method.

```
public int compute(int n, int k) {
int answer = 1;
for (int i = 1; i <= k; i++)
answer *= n;
return answer;
}
```

Which of the following represents the value returned as a result of the call compute([tex]$n, k$[/tex])?

A. [tex]n^k[/tex]

B. [tex]n![/tex]

C. [tex]2^k[/tex]

D. [tex]k^n[/tex]



Answer :

To determine the value returned as the result of the call `compute(n, k)`, let's analyze the given method step-by-step:

1. The method initializes an integer variable `answer` to 1.

```java
int answer = 1;
```

2. It then uses a `for` loop to iterate `k` times. For each iteration, the variable `answer` is multiplied by `n`.

```java
for (int i = 1; i <= k; i++)
answer = n;
```

3. Finally, the method returns the value of `answer`.

```java
return answer;
```

### Detailed Steps

- The loop starts with `answer` as 1.
- In each iteration, `answer` is multiplied by `n`.
- This multiplication happens `k` times.

Effectively, what this series of multiplications is doing is raising `n` to the power of `k`. Here’s why:

- After the first iteration, `answer` = [tex]\( n \)[/tex]
- After the second iteration, `answer` = [tex]\( n \times n = n^2 \)[/tex]
- After the third iteration, `answer` = [tex]\( n \times n \times n = n^3 \)[/tex]
- And so on…

By the end of `k` iterations, `answer` will be [tex]\( n^k \)[/tex].

Therefore, the value returned by the method `compute(n, k)` is [tex]\( n^k \)[/tex].

Given the options:

- [tex]\( n^
k \)[/tex]
- [tex]\( n! \)[/tex]
- [tex]\( n \% 5 Ek'' \)[/tex]
- [tex]\( 2 \% 5 E k' \)[/tex]
- [tex]\( k^n \)[/tex]
- [tex]\( n^k \)[/tex]

The correct answer representing the value returned as a result of the call `compute(n, k)` is:

[tex]\( n^k \)[/tex].