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

Explanation

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

### Step 1: Identify 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 \( n = 50 \).

### 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 Two Terms Equal
Since we want the 17th and 18th terms to be equal:
\[ T_{17} = T_{18} \]
This gives us:
\[ \binom{50}{16} (2)^{34} (a)^{16} = \binom{50}{17} (2)^{33} (a)^{17} \]

### Step 4: Simplify the Equation
We can simplify this equation. First, note that:
\[ \binom{50}{17} = \frac{50!}{17! \cdot 33!} \quad \text{and} \quad \binom{50}{16} = \frac{50!}{16! \cdot 34!} \]
Thus:
\[ \frac{\binom{50}{16}}{\binom{50}{17}} = \frac{34}{17} = 2 \]

Substituting this back into our equation:
\[ 2 \cdot (2)^{34} (a)^{16} = (2)^{33} (a)^{17} \]

### Step 5: Further Simplify
Now we can simplify:
\[ 2^{35} (a)^{16} = 2^{33} (a)^{17} \]
Dividing both sides by \( 2^{33} \):
\[ 2^{2} (a)^{16} = (a)^{17} \]
This simplifies to:
\[ 4 = a \]

### Step 6: Solve for \( a \)
Thus, we find:
\[ a = \frac{4}{1} = 4 \]

### Final Answer
The value of \( a \) is:
\[ \boxed{4} \]