Sequences & Series
Summation using Binomial Coefficients
Grade 11
Question:
<p>If \(S_n = \displaystyle\sum_{r=0}^{n} \dfrac{1}{{}^nC_r}\) and \(t_n = \displaystyle\sum_{r=0}^{n} \dfrac{r}{{}^nC_r}\), then \(\dfrac{t_n}{S_n}\) equals:</p>
<p>\(n\)</p>
<p>\(\dfrac{n}{2}\)</p>
<p>\(\dfrac{n}{2} - 1\)</p>
<p>\(2n\)</p>
Step-by-Step Solution
Key Concept: Use the symmetry property of binomial coefficients: $\binom{n}{r} = \binom{n}{n-r}$. When you pair terms in $t_n$ by replacing $r$ with $n-r$, the numerator becomes $n-r$ while the denominator stays the same, allowing you to express $t_n$ in terms of $S_n$.
<p><strong>Step 1:</strong> Write out $t_n$ using the symmetry property. For each term $\frac{r}{\binom{n}{r}}$, pair it with the term $\frac{n-r}{\binom{n}{n-r}}$.</p><p><strong>Step 2:</strong> Since $\binom{n}{r} = \binom{n}{n-r}$, we have: $$t_n = \sum_{r=0}^{n} \frac{r}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n-r}{\binom{n}{r}}$$</p><p><strong>Step 3:</strong> Add these two expressions for $t_n$: $$2t_n = \sum_{r=0}^{n} \frac{r + (n-r)}{\binom{n}{r}} = \sum_{r=0}^{n} \frac{n}{\binom{n}{r}} = n\sum_{r=0}^{n} \frac{1}{\binom{n}{r}} = nS_n$$</p><p><strong>Step 4:</strong> Therefore: $$\frac{t_n}{S_n} = \frac{n}{2}$$</p><p>∴ Answer: B</p>
Correct Answer: B