A function is defined as follows :
`f(a_(1), a_(2), a_(3)...a_(n))=a_(1)2^(n-1)+a_(2)2^(n-2)+a_(3)2^(n-3)+...a_(n)2^(0)`
The above function is repreated until the value of function reduces to a single digit number.
`f(128)` equals :

Options

A. 1
B. 2
C. 4
D. 8

Answer

To solve the problem, we need to evaluate the function \( f(a_1, a_2, a_3, \ldots, a_n) \) defined as:

\[ f(a_1, a_2, a_3, \ldots, a_n) = a_1 \cdot 2^{n-1} + a_2 \cdot 2^{n-2} + a_3 \cdot 2^{n-3} + \ldots + a_n \cdot 2^{0} \]

We will apply this function repeatedly until we reach a single-digit number, starting with \( f(128) \).

### Step 1: Identify the digits of 128
The number 128 has three digits: \( a_1 = 1 \), \( a_2 = 2 \), and \( a_3 = 8 \). Thus, we can represent it as \( f(1, 2, 8) \).

### Step 2: Calculate \( f(1, 2, 8) \)
Using the function definition:
- Here, \( n = 3 \) (since there are three digits).
- We calculate:

\[ f(1, 2, 8) = 1 \cdot 2^{3-1} + 2 \cdot 2^{3-2} + 8 \cdot 2^{3-3} \]

Calculating each term:
- \( 1 \cdot 2^{2} = 1 \cdot 4 = 4 \)
- \( 2 \cdot 2^{1} = 2 \cdot 2 = 4 \)
- \( 8 \cdot 2^{0} = 8 \cdot 1 = 8 \)

Now, summing these values:
\[ f(1, 2, 8) = 4 + 4 + 8 = 16 \]

### Step 3: Repeat the function with the result
Now we need to evaluate \( f(16) \). The number 16 has two digits: \( a_1 = 1 \) and \( a_2 = 6 \).

### Step 4: Calculate \( f(1, 6) \)
Using the function definition again:
- Here, \( n = 2 \) (since there are two digits).
- We calculate:

\[ f(1, 6) = 1 \cdot 2^{2-1} + 6 \cdot 2^{2-2} \]

Calculating each term:
- \( 1 \cdot 2^{1} = 1 \cdot 2 = 2 \)
- \( 6 \cdot 2^{0} = 6 \cdot 1 = 6 \)

Now, summing these values:
\[ f(1, 6) = 2 + 6 = 8 \]

### Step 5: Check if the result is a single digit
The result is 8, which is a single-digit number. Therefore, we stop here.

### Final Answer
The final answer is:
\[ f(128) = 8 \]

---