Binomial Theorem
Properties of Binomial Coefficients
Grade 11
Question:
<p>If \(a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}}\), then \(\sum_{r=0}^{n} \frac{r}{\binom{n}{r}}\) is equal to</p>
<p>(a) \((n-1)a_n\)</p>
<p>(b) \(na_n\)</p>
<p>(c) \(\frac{1}{n}a_n\)</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the property that $\frac{r}{\binom{n}{r}} = n \cdot \frac{1}{\binom{n-1}{r-1}}$ to relate the sum to the definition of $a_n$.
<p><strong>Solution:</strong> We have $a_n = \sum_{r=0}^{n} \frac{1}{\binom{n}{r}}$</p><p>We need to find $\sum_{r=0}^{n} \frac{r}{\binom{n}{r}}$</p><p>Note that $\frac{r}{\binom{n}{r}} = \frac{r \cdot n!}{r! (n-r)!} = \frac{n!}{(r-1)!(n-r)!} = \frac{n \cdot (n-1)!}{(r-1)!((n-1)-(r-1))!} = n \cdot \frac{1}{\binom{n-1}{r-1}}$</p><p>Therefore, $\sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = n \sum_{r=1}^{n} \frac{1}{\binom{n-1}{r-1}} = n \sum_{s=0}^{n-1} \frac{1}{\binom{n-1}{s}} = n \cdot a_{n-1} = (n-1)a_n$</p><p>∴ Answer is (a).</p>
Correct Answer: A