Question

If \( 0<x<1 \), then first negative term in the expansion of \( (1+x)^{\frac{47}{5}} \) is

• A. \( 10^{\text{th}} \) term
• B. \( 11^{\text{th}} \) term
• C. \( 12^{\text{th}} \) term
• D. \( 13^{\text{th}} \) term

Hint:

For any positive fractional exponent \( n \), the number of positive terms before the negative terms start is equal to \( \lfloor n \rfloor + 1 \). Here, \( \lfloor 9.4 \rfloor + 1 = 10 \) positive terms, so the 11th term must be the first negative one!

Answer 1

The Correct Option is C

Step 1: Recall the general term.
For $(1+x)^n$ the term is \[ T_{r+1} = \frac{n(n-1)\cdots(n-r+1)}{r!}x^r \] Since $x>0$, only the top product decides the sign.
Step 2: Find when a term turns negative.
The top stays positive while $n-r+1 > 0$. The first negative term shows up when $n-r+1 < 0$, that is $r > n+1$.
Step 3: Put in the value of n.
Here $n = \dfrac{47}{5} = 9.4$. So we need $r > 9.4 + 1 = 10.4$.
Step 4: Pick the smallest whole r.
The smallest integer above $10.4$ is $r = 11$.
Step 5: Convert r to a term number.
The term is $T_{r+1}$, so the position is $11+1 = 12$.
Step 6: State the answer.
The first negative term is the \[ \boxed{12\text{th term}} \]