To resolve this problem, the binomial theorem is employed to determine expansion terms, which are then utilized to find the coefficient in another expansion.
The given expression is \((2+a)^{50}\), and the equality of its 17th and 18th terms is the focus.
The general term in the binomial expansion of \((x+y)^n\) is defined as:
\(T_{k+1} = \binom{n}{k}x^{n-k}y^k\)
For the expression \((2+a)^{50}\), the 17th term is calculated as:
\(T_{17} = \binom{50}{16}(2)^{50-16}a^{16} = \binom{50}{16}(2)^{34}a^{16}\)
The 18th term is:
\(T_{18} = \binom{50}{17}(2)^{50-17}a^{17} = \binom{50}{17}(2)^{33}a^{17}\)
The problem states that these two terms are equal:
\(\binom{50}{16}(2)^{34}a^{16} = \binom{50}{17}(2)^{33}a^{17}\)
Upon dividing both sides by \(\binom{50}{17}(2)^{33}a^{16}\), we obtain:
\(\frac{\binom{50}{16}(2)^{34}}{\binom{50}{17}(2)^{33}} = a\)
Simplification yields:
\(\frac{\binom{50}{16} \cdot 2}{\binom{50}{17}} = a\)
Using the identity \(\binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1}\), specifically \(\binom{50}{16} = \frac{50-16+1}{16}\cdot \binom{50}{17} = \frac{35}{16}\cdot \binom{50}{17}\). Wait, there was a mistake in calculation. The correct identity is \(\binom{n}{k} = \frac{n}{k} \binom{n-1}{k-1}\) and \(\binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1}\). Let's use \(\binom{n}{k} = \frac{n-k+1}{k} \binom{n}{k-1}\). So \(\binom{50}{16} = \frac{50-16+1}{16} \binom{50}{17}\) is incorrect. It should be \(\binom{50}{17} = \frac{50-17+1}{17} \binom{50}{16} = \frac{34}{17} \binom{50}{16} = 2 \binom{50}{16}\), which means \(\binom{50}{16} = \frac{1}{2} \binom{50}{17}\). The provided identity was \(\binom{50}{16} = \frac{50-16+1}{16}\cdot \binom{50}{17}\) which implies \(\binom{50}{16} = \frac{35}{16}\binom{50}{17}\). This is incorrect. Let's re-evaluate the original approach where \(\frac{\binom{50}{16}}{\binom{50}{17}} = \frac{17}{50-17+1} = \frac{17}{34} = \frac{1}{2}\). So, \(a = \frac{1}{2} \cdot 2 = 1\).
Now, we use the value \(a=1\) for the expansion of \((a+x)^{-2}\), specifically the coefficient of \(x^{35}\). The expression is \((1+x)^{-2}\).
The general term for \((a+x)^n\) is:
\(T_{k+1} = \binom{n}{k}a^{n-k}x^k\)
For \((1+x)^{-2}\), the general term is:
\(T_{k+1} = \binom{-2}{k}(1)^{-2-k}x^k = \binom{-2}{k}x^k\)
Using the formula for negative indices: \(\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}\)
We need the term where \(k=35\), which is \(T_{36}\):
\(T_{36} = \binom{-2}{35}(1)^{-2-35}x^{35} = \binom{-2}{35}x^{35}\)
Calculating the binomial coefficient:
\(\binom{-2}{35} = (-1)^{35}\binom{2+35-1}{35} = (-1)^{35}\binom{36}{35}\)
Since \(\binom{36}{35} = \binom{36}{36-35} = \binom{36}{1} = 36\):
\(\binom{-2}{35} = (-1)^{35} \cdot 36 = -36\)
Thus, the coefficient of \(x^{35}\) is \(-36\).