Binomial Theorem
Independent of Variable Term
Grade 11
Question:
<p>The term independent of \(x\) in \((1+x)^m \left(1 + \frac{1}{x}\right)^n\) is</p>
<p>(a) \(\binom{m+n}{m}\)</p>
<p>(b) \(\binom{m+n}{n+1}\)</p>
<p>(c) \(\binom{m+n}{m-n}\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Identify that the constant term requires $p = q$, then sum over all valid $p$ using the Vandermonde or Chu-Vandermonde identity.
<p><strong>Solution:</strong> The general term from $(1+x)^m$ is $\binom{m}{p} x^p$ and from $\left(1 + \frac{1}{x}\right)^n$ is $\binom{n}{q} x^{-q}$.</p><p>The combined term is $\binom{m}{p}\binom{n}{q} x^{p-q}$.</p><p>For independence from $x$: $p - q = 0 \implies p = q$.</p><p>The constant term is $\sum_{p=0}^{\min(m,n)} \binom{m}{p}\binom{n}{p}$.</p><p>By Vandermonde's identity variant: this equals $\binom{m+n}{m}$.</p>
Correct Answer: A