`sum_(r=0)^(4) (-1)^(r )""^(16)C_(r)` is divisible by :

Options

A. 5
B. 7
C. 11
D. 13

Answer

To solve the problem, we need to evaluate the summation:

\[ \sum_{r=0}^{4} (-1)^{r} \binom{16}{r} \]

### Step 1: Write out the terms of the summation

The summation can be expanded as follows:

\[ \binom{16}{0} (-1)^0 + \binom{16}{1} (-1)^1 + \binom{16}{2} (-1)^2 + \binom{16}{3} (-1)^3 + \binom{16}{4} (-1)^4 \]

This simplifies to:

\[ \binom{16}{0} - \binom{16}{1} + \binom{16}{2} - \binom{16}{3} + \binom{16}{4} \]

### Step 2: Calculate the binomial coefficients

Now we calculate each of the binomial coefficients:

- \(\binom{16}{0} = 1\)
- \(\binom{16}{1} = 16\)
- \(\binom{16}{2} = \frac{16 \times 15}{2} = 120\)
- \(\binom{16}{3} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 560\)
- \(\binom{16}{4} = \frac{16 \times 15 \times 14 \times 13}{4 \times 3 \times 2 \times 1} = 1820\)

### Step 3: Substitute the values back into the summation

Now substituting these values back into the summation:

\[ 1 - 16 + 120 - 560 + 1820 \]

### Step 4: Simplify the expression

Now, we will perform the calculations step by step:

1. \(1 - 16 = -15\)
2. \(-15 + 120 = 105\)
3. \(105 - 560 = -455\)
4. \(-455 + 1820 = 1365\)

Thus, the value of the summation is:

\[ \sum_{r=0}^{4} (-1)^{r} \binom{16}{r} = 1365 \]

### Step 5: Check for divisibility

Now we need to check if 1365 is divisible by the given numbers: 5, 7, 11, and 13.

1. **Divisibility by 5**:
- The last digit of 1365 is 5, so it is divisible by 5.

2. **Divisibility by 7**:
- \(1365 \div 7 = 195\) remainder \(0\) (not divisible).

3. **Divisibility by 11**:
- \(1365 \div 11 = 124.09\) (not divisible).

4. **Divisibility by 13**:
- \(1365 \div 13 = 105\) remainder \(0\) (divisible).

### Conclusion

The final answer is that the sum \( \sum_{r=0}^{4} (-1)^{r} \binom{16}{r} \) is divisible by **5** and **13**.