Question

If the 17th and the 18th terms in the expansion of \( (2 + a)^{50} \) are equal, then the coefficient of \( x^{35} \) in the expansion of \( (a + x)^{-2} \) is:

• A. -35
• B. 3
• C. 36
• D. -36

Hint:

For \( (1+x)^{-2} \), the expansion is \( 1 - 2x + 3x^2 - 4x^3 + \dots \). The coefficient of \( x^n \) is always \( (-1)^n (n+1) \).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
First, we find the value of \( a \) by equating the general terms of the binomial expansion. Then, we use the general term formula for a negative binomial expansion to find the required coefficient.

Step 2: Key Formula or Approach:
1. General term \( T_{r+1} \) for \( (x+y)^n \) is \( \binom{n}{r} x^{n-r} y^r \).
2. General term of \( (1+z)^{-n} \) is \( (-1)^r \binom{n+r-1}{r} z^r \).

Step 3: Detailed Explanation:
1. Find \( a \):
Given \( T_{17} = T_{18} \) in \( (2+a)^{50} \).
\[ \binom{50}{16} 2^{34} a^{16} = \binom{50}{17} 2^{33} a^{17} \] \[ \frac{\binom{50}{16}}{\binom{50}{17}} \times \frac{2^{34}}{2^{33}} = \frac{a^{17}}{a^{16}} \] \[ \frac{17}{50-17+1} \times 2 = a \implies \frac{17}{34} \times 2 = a \implies a = 1 \] 2. Find coefficient in \( (1+x)^{-2} \):
Since \( a = 1 \), the expression is \( (1+x)^{-2} \).
The general term is \( T_{r+1} = (-1)^r (r+1) x^r \).
For \( x^{35} \), \( r = 35 \):
\[ \text{Coefficient} = (-1)^{35} (35 + 1) = -1 \times 36 = -36 \]

Step 4: Final Answer
The coefficient of \( x^{35} \) is \( -36 \).