Find a if 17th and 18th terms in the expansion of `(2+a)^(50)` are equal.

Options

A. `1//3`
B. `1//2`
C. `1`
D. None of these

Explanation

To find the value of \( a \) such that the 17th and 18th terms in the expansion of \( (2 + a)^{50} \) are equal, we can follow these steps:

### Step 1: Write the General Term
The general term \( T_{r+1} \) in the binomial expansion of \( (P + Q)^N \) is given by:
\[ T_{r+1} = \binom{N}{r} P^{N-r} Q^r \]
For our case, \( P = 2 \), \( Q = a \), and \( N = 50 \). Thus, the general term becomes:
\[ T_{r+1} = \binom{50}{r} (2)^{50-r} (a)^r \]

### Step 2: Write the 17th and 18th Terms
The 17th term \( T_{17} \) corresponds to \( r = 16 \):
\[ T_{17} = \binom{50}{16} (2)^{50-16} (a)^{16} = \binom{50}{16} (2)^{34} (a)^{16} \]
The 18th term \( T_{18} \) corresponds to \( r = 17 \):
\[ T_{18} = \binom{50}{17} (2)^{50-17} (a)^{17} = \binom{50}{17} (2)^{33} (a)^{17} \]

### Step 3: Set the 17th and 18th Terms Equal
Since we want \( T_{17} = T_{18} \):
\[ \binom{50}{16} (2)^{34} (a)^{16} = \binom{50}{17} (2)^{33} (a)^{17} \]

### Step 4: Simplify the Equation
We can divide both sides by \( (2)^{33} \):
\[ \binom{50}{16} (2) (a)^{16} = \binom{50}{17} (a)^{17} \]
Now, we can express \( \binom{50}{17} \) in terms of \( \binom{50}{16} \):
\[ \binom{50}{17} = \frac{50 - 17 + 1}{17} \binom{50}{16} = \frac{34}{17} \binom{50}{16} \]
Substituting this back into the equation gives:
\[ 2 (a)^{16} \binom{50}{16} = \frac{34}{17} \binom{50}{16} (a)^{17} \]

### Step 5: Cancel \( \binom{50}{16} \) and Rearrange
Assuming \( \binom{50}{16} \neq 0 \), we can cancel it:
\[ 2 (a)^{16} = \frac{34}{17} (a)^{17} \]
Rearranging gives:
\[ 2 = \frac{34}{17} a \]

### Step 6: Solve for \( a \)
Multiplying both sides by \( 17 \):
\[ 34 = 34a \]
Dividing both sides by \( 34 \):
\[ a = 1 \]

### Final Answer
Thus, the value of \( a \) is:
\[ \boxed{1} \]