Question

If \( x \) is a positive real number and the first negative term in the expansion of \[ (1 + x)^{27/5} \text{ is } t_k, \text{ then } k =\ ? \]

• A. \( 5 \)
• B. \( 6 \)
• C. \( 7 \)
• D. \( 8 \)

Hint:

When expanding a binomial with fractional powers, the first negative term occurs when the numerator in the binomial coefficient becomes negative.

Answer 1

The Correct Option is D

Step 1: Consider the general term in the expansion of \( (1 + x)^n \) when \( n \) is a rational number: \[ t_{k+1} = \binom{n}{k} x^k \] Here, \( n = \frac{27}{5} \), so: \[ t_{k+1} = \binom{27/5}{k} x^k \] Step 2: The term becomes negative when \( \binom{27/5}{k} \) is negative. Since \( \binom{n}{k} = \frac{n(n - 1)(n - 2).s(n - k + 1)}{k!} \), the first negative binomial coefficient appears when \( n - (k - 1)<0 \), i.e., \[ \frac{27}{5} - (k - 1)<0 \Rightarrow k>\frac{32}{5} = 6.4 \] Step 3: The smallest integer \( k \) satisfying this is \( 7 \), so the first negative term is: \[ t_{k+1} = t_8 \] Thus, \( k = 8 \)