Binomial Theorem
Grade None
Question:
<p>If x is positive, the first negative term in the expansion of <span class="math-tex">\((1+x)^{\frac{27}{5}}\)</span> is</p>
<p style="display:inline">8<sup>th</sup> term</p>
<p style="display:inline">5<sup>th</sup> term</p>
<p style="display:inline">7<sup>th</sup> term</p>
<p style="display:inline">6<sup>th</sup> term</p>
Step-by-Step Solution
Key Concept: The first negative term in the expansion of $(1+x)^n$ for $x > 0$ is determined by finding the smallest integer $r$ such that the factor $(n - r + 1)$ in the general term's numerator becomes negative.
<p>T<sub>r+1</sub> = <span class="math-tex">$\frac{n(n-1)(n-2) \ldots(n-r+1)}{r !}$</span> (x)<sup>r</sup><br />
For first negative term n - r + 1 < 0<br />
i.r., <span class="math-tex">$\frac{27}{5}$</span> - r + 1 < 0 <span class="math-tex">$\Leftrightarrow$</span> r > <span class="math-tex">$\frac{32}{5} \Leftrightarrow$</span> r = 7<br />
The first negative term is (7 + 1)<sup>th</sup> i.e., 8<sup>th</sup> term.</p>
Correct Answer: A