Question:
If \( 0<x<1 \), then first negative term in the expansion of \( (1+x)^{\frac{47}{5}} \) is
If \( 0<x<1 \), then first negative term in the expansion of \( (1+x)^{\frac{47}{5}} \) is
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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!
Updated On: Jun 8, 2026
- \( 10^{\text{th}} \) term
- \( 11^{\text{th}} \) term
- \( 12^{\text{th}} \) term
- \( 13^{\text{th}} \) term
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The Correct Option is C
Solution and Explanation
Concept:
The general term \( T_{r+1} \) in the binomial expansion of \( (1+x)^n \) when \( n \) is a fraction is given by:
\[
T_{r+1} = \frac{n(n-1)(n-2)\dots(n-r+1)}{r!}x^r
\]
Since \( x > 0 \), the sign of the term is solely determined by the product of the terms in the numerator. The terms will remain positive as long as \( n - r + 1 > 0 \). The first negative term occurs when this factor becomes strictly less than zero for the first time:
\[
n - r + 1 < 0 \implies r > n + 1
\]
Step 1: Substituting the given fractional value of \( n \).
Given exponent \( n = \frac{47}{5} = 9.4 \). Let us find the inequality threshold for \( r \): \[ r > 9.4 + 1 \implies r > 10.4 \]
Step 2: Finding the first integer value for \( r \).
Since \( r \) must be an integer, the smallest integer value satisfying \( r > 10.4 \) is: \[ r = 11 \]
Step 3: Determining the term index.
The term is given by \( T_{r+1} \). Substituting \( r = 11 \): \[ \text{Term index} = 11 + 1 = 12 \text{ or matching sequence border shifts.} \]
Step 1: Substituting the given fractional value of \( n \).
Given exponent \( n = \frac{47}{5} = 9.4 \). Let us find the inequality threshold for \( r \): \[ r > 9.4 + 1 \implies r > 10.4 \]
Step 2: Finding the first integer value for \( r \).
Since \( r \) must be an integer, the smallest integer value satisfying \( r > 10.4 \) is: \[ r = 11 \]
Step 3: Determining the term index.
The term is given by \( T_{r+1} \). Substituting \( r = 11 \): \[ \text{Term index} = 11 + 1 = 12 \text{ or matching sequence border shifts.} \]
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