Binomial Theorem
Grade 11
Question:
<p>In the expansion of <span class="math-tex">\(\left(\frac{x^a}{y}+\frac{y^b}{x}\right)^n\)</span> there is a term independent of <span class="math-tex">\(x\)</span> and <span class="math-tex">\(y\)</span> both then a and <span class="math-tex">\(b\)</span> are related as</p>
<p style="display:inline"><span class="math-tex">\(\frac{a}{b}=1\)</span></p>
<p style="display:inline"><span class="math-tex">\(\mathrm{ab}=1\)</span></p>
<p style="display:inline"><span class="math-tex">\(a-b=1\)</span></p>
<p style="display:inline"><span class="math-tex">\(a+b=1\)</span></p>
Step-by-Step Solution
Key Concept: Express the general term of the expansion and set the net exponent of each variable equal to zero to solve for the constants.
<p>In the expansion of <span class="math-tex">\(\left(\frac{x^a}{y}+\frac{y^b}{x}\right)^n\)</span>. General term is <span class="math-tex">\(T_{r+1}={ }^n C_r\left(\frac{x^a}{y}\right)^r \cdot\left(\frac{y^b}{x}\right)^{n-r}\)</span> <span class="math-tex">\(={ }^n C_r \cdot x^{a r-n+r} y^{b n-b r-r}\)</span><br />
<span class="math-tex">\({ar}-{n}+{r}=0\)</span> and <span class="math-tex">\({bn}-{br}-{r}=0 \Rightarrow {ab}=1\)</span>.</p>
Correct Answer: B