If `17^(th)` and `18^(th)` terms in the expension of `(2+a)^(50)` are equal ,then the value of a is :

Options

A. 0
B. `-1`
C. 2
D. 1

Answer

To solve the problem, we need to find the value of \( a \) such that the 17th and 18th terms in the expansion of \( (2 + a)^{50} \) are equal.

### Step 1: Write the general term of the binomial expansion
The general term \( T_k \) in the expansion of \( (x + y)^n \) is given by:
\[ T_k = \binom{n}{k-1} x^{n-(k-1)} y^{k-1} \]
In our case, \( x = 2 \), \( y = a \), and \( n = 50 \).

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

### Step 3: Set the two terms equal
Since \( T_{17} = T_{18} \), we have:
\[ \binom{50}{16} (2)^{34} (a)^{16} = \binom{50}{17} (2)^{33} (a)^{17} \]

### Step 4: Simplify the equation
We can simplify this equation. First, divide both sides by \( (2)^{33} \):
\[ \binom{50}{16} (2) (a)^{16} = \binom{50}{17} (a)^{17} \]
Now, divide both sides by \( (a)^{16} \) (assuming \( a \neq 0 \)):
\[ \binom{50}{16} (2) = \binom{50}{17} (a) \]

### Step 5: Use the property of binomial coefficients
Recall that:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Thus, we can express \( \binom{50}{17} \) in terms of \( \binom{50}{16} \):
\[ \binom{50}{17} = \frac{50 - 17 + 1}{17} \binom{50}{16} = \frac{34}{17} \binom{50}{16} = 2 \binom{50}{16} \]

### Step 6: Substitute back into the equation
Now substituting back, we get:
\[ \binom{50}{16} (2) = 2 \binom{50}{16} (a) \]
Dividing both sides by \( \binom{50}{16} \) (assuming it is not zero):
\[ 2 = 2a \]

### Step 7: Solve for \( a \)
Now, divide both sides by 2:
\[ 1 = a \]

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