To solve the problem, we need to address it in two parts. First, find the value of \( a \) from the binomial expansion condition, then use that \( a \) value to determine the coefficient of \( x^{35} \) in the expansion of \( (a + x)^{-2} \).
- Finding \( a \) from the binomial expansion condition:
- The binomial expansion of \((2 + a)^{50}\) has the general term \( T_r = \binom{50}{r} \cdot 2^{50-r} \cdot a^r \).
- The 17th term, \( T_{17} \), corresponds to \( r = 16 \): \( T_{17} = \binom{50}{16} \cdot 2^{34} \cdot a^{16} \).
- The 18th term, \( T_{18} \), corresponds to \( r = 17 \): \( T_{18} = \binom{50}{17} \cdot 2^{33} \cdot a^{17} \).
- According to the problem, \( T_{17} = T_{18} \).
- Set the terms equal: \(\binom{50}{16} \cdot 2^{34} \cdot a^{16} = \binom{50}{17} \cdot 2^{33} \cdot a^{17}\).
- Cancel common terms and simplify: \(2a = \frac{\binom{50}{17}}{\binom{50}{16}}\).
- Simplifying binomial coefficients: \(\frac{\binom{50}{17}}{\binom{50}{16}} = \frac{50 - 16}{17} = \frac{34}{17} = 2\).
- This implies \( 2a = 2 \), hence \( a = 1 \).
- Finding the coefficient of \( x^{35} \) in the expansion of \((a + x)^{-2}\) with \( a = 1\):
- The binomial expansion of \((1 + x)^{-2}\) has a general term for negative binomials: \(T_k = \binom{-2}{k} \cdot 1^{n-k} \cdot x^k\).
- Using the formula for negative binomial coefficients: \(\binom{-2}{k} = \frac{(-2)(-3)\cdots(-2-k+1)}{k!} = \frac{(-1)^k \cdot (k+1)}{k!}\).
- We want the term where \( k = 35 \): \(T_{35} = \binom{-2}{35} \cdot x^{35}\).
- Calculate \(\binom{-2}{35}\): \(\binom{-2}{35} = \frac{(-1)^{35} \cdot (35+1)}{35!} = \frac{-36}{35!}\).
- Therefore, the coefficient of \( x^{35} \) is \(-36\).
Thus, the coefficient of \( x^{35} \) in the expansion of \( (1 + x)^{-2} \) is -36.