`AA n in N, 49^n+16n-1` is divisible by (A) `64` (B) `49` (C) `132` (D) `32`

Options

A. 64
B. 8
C. 16
D. 4

Answer

To determine the divisibility of the expression \(49^n + 16n - 1\) by the given options, we will use the binomial theorem and some algebraic manipulation.

### Step-by-Step Solution:

1. **Rewrite the expression**:
We start with the expression \(49^n + 16n - 1\). Notice that \(49\) can be rewritten as \(1 + 48\). Therefore, we can express \(49^n\) as:
\[ 49^n = (1 + 48)^n \]

2. **Apply the Binomial Theorem**:
According to the binomial theorem, we can expand \((1 + 48)^n\):
\[ (1 + 48)^n = \sum_{k=0}^{n} \binom{n}{k} 48^k \]
This expands to:
\[ 1 + \binom{n}{1} 48 + \binom{n}{2} 48^2 + \binom{n}{3} 48^3 + \ldots + 48^n \]

3. **Combine with \(16n - 1\)**:
Now, substituting back into our expression:
\[ 49^n + 16n - 1 = \left(1 + \binom{n}{1} 48 + \binom{n}{2} 48^2 + \ldots + 48^n\right) + 16n - 1 \]
The \(1\) and \(-1\) cancel out, leaving us with:
\[ \binom{n}{1} 48 + \binom{n}{2} 48^2 + \ldots + 48^n + 16n \]

4. **Factor out common terms**:
We can factor out \(16\) from \(16n\) and observe that \(48\) can be expressed as \(3 \times 16\):
\[ = 16 \left(\frac{n}{3} + \text{other terms involving } 48\right) \]

5. **Check divisibility by \(64\)**:
Notice that \(48\) is divisible by \(16\), and since we have multiple terms involving \(48\), we can check how many factors of \(16\) we can extract from the entire expression. Each term contributes at least one factor of \(16\), and since \(48\) is \(3 \times 16\), we can conclude that:
\[ 49^n + 16n - 1 \text{ is divisible by } 64 \]

### Conclusion:
Thus, the expression \(49^n + 16n - 1\) is divisible by \(64\). Therefore, the correct answer is **(A) 64**.