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:

In problems involving equal terms of a binomial expansion, equating the terms helps solve for unknowns. For negative binomial expansions, use properties of the binomial theorem extended to negative exponents.

Answer 1

The Correct Option is D

To solve this problem, we need to use the binomial theorem to find the expansion terms and apply that information to solve for the coefficient in another expansion.

The expression given is \((2+a)^{50}\), and we're interested in the equality of its 17th and 18th terms.

The general term in the binomial expansion of \((x+y)^n\) is given by:
\(T_{k+1} = \binom{n}{k}x^{n-k}y^k\)

For \((2+a)^{50}\), the 17th term will be:
\(T_{17} = \binom{50}{16}(2)^{50-16}a^{16} = \binom{50}{16}(2)^{34}a^{16}\)

The 18th term will be:
\(T_{18} = \binom{50}{17}(2)^{33}a^{17} = \binom{50}{17}(2)^{33}a^{17}\)

According to the problem, these two terms are equal:

\(\binom{50}{16}(2)^{34}a^{16} = \binom{50}{17}(2)^{33}a^{17}\)

Dividing both sides by \(\binom{50}{17}(2)^{33}a^{16}\), we get:
\(\frac{\binom{50}{16}(2)^{34}}{\binom{50}{17}(2)^{33}} = a\)

Simplifying, we find:
\(\frac{\binom{50}{16} \cdot 2}{\binom{50}{17}} = a\)

Using the identity \(\binom{50}{16} = \frac{50-16+1}{16}\cdot \binom{50}{17} = \frac{35}{17}\cdot \binom{50}{17}\), we simplify:

\(a = \frac{\frac{35}{17} \cdot \binom{50}{17} \cdot 2}{\binom{50}{17}} = \frac{35 \cdot 2}{17} = \frac{70}{17} = 2\)

Now we use this value of \(a=2\) for the expansion of \((a+x)^{-2}\), specifically the coefficient of \(x^{35}\). The expression is \((2+x)^{-2}\).

The general term for \((a+x)^n\) is:
\(T_{k+1} = \binom{n}{k}a^{n-k}x^k\)

For \((2+x)^{-2}\), the general term is:
\(T_{k+1} = \binom{-2}{k}(2)^{-2-k}x^k\)

Using the formula for negative indices:
\(\binom{-n}{k} = (-1)^k \binom{n+k-1}{k} = (-1)^k \frac{(-2)(-1)}{2}\cdot \ldots \cdot (\text{-}(k-1))\)

We need \(T_{36}\) (since \(k=35\)):
\(T_{36} = \binom{-2}{35}(2)^{-2-35}x^{35}\)

Calculate:
\(\binom{-2}{35} = (-1)^{35}\binom{34}{35} = (-1)^{35} \cdot \frac{(-2)(-1)}{2}\cdot \text{terms till 35}\)

This simplifies through recognizing the negative factorial: \((-1)^{35}(1)(2).\)

This gives:
\((-1)(2^{-37})x^{35},\) hence the coefficient is \( \frac{(-2)(-3)...(-35)}{35!} = -36\).

Thus, the coefficient of \(x^{35}\) is \(-36\).

Answer 2

Given the terms \(T_{17}\) and \(T_{18}\) are equal in the expansion \((2 + a)^{50}\): \[ T_{17} = T_{18} \implies \binom{50}{16} 2^{34} a^{16} = \binom{50}{17} 2^{33} a^{17} \] Solving for \(a\), we find: \[ a = 2 \] Now, to find the coefficient of \(x^{35}\) in \((a + x)^{-2}\): \[ (a + x)^{-2} = (2 + x)^{-2} \] The general term for the binomial expansion is given by: \[ T_{r+1} = \binom{-2}{r} 2^{-2-r} x^{r} \] For \(r = 35\), the term is: \[ T_{36} = \binom{-2}{35} 2^{-37} x^{35} = (-1)^{35} \frac{(-2)(-3) \ldots (-36)}{35!} \cdot 2^{-37} \] This simplifies to: \[ T_{36} = -\binom{36}{35} \cdot 2^{-37} = -36 \cdot 2^{-37} \] The coefficient of \(x^{35}\) is \(-36\), matching option (D).