Question

If $a_n = \sum_{r=0}^n \frac{1}{^{n}C_r}$ and $b_n = \sum_{r=0}^n \frac{r}{^{n}C_r}$, then $\frac{b_n}{a_n} =$

• A. $n$
• B. $\frac{n}{2}$
• C. $\frac{n}{3}$
• D. $n - 1$

Hint:

For a quick test, substitute a small value like $n = 1$. This gives $a_1 = \frac{1}{1} + \frac{1}{1} = 2$ and $b_1 = \frac{0}{1} + \frac{1}{1} = 1$. The ratio is $\frac{1}{2}$, which matches $\frac{n}{2}$.

Answer 1

The Correct Option is B

Step 1: Concept
We can rewrite the summation for $b_n$ by using the property of binomial coefficients: $^{n}C_r = ^{n}C_{n-r}$.

Step 2: Meaning
Reversing the order of summation terms allows us to express $b_n$ in terms of $a_n$ and solve for their ratio.

Step 3: Analysis
Substitute $r = n-r$ in the expression for $b_n$: \[ b_n = \sum_{r=0}^n \frac{r}{^{n}C_r} = \sum_{r=0}^n \frac{n-r}{^{n}C_{n-r}} = \sum_{r=0}^n \frac{n-r}{^{n}C_r} \] Adding the two expressions for $b_n$ yields: \[ 2b_n = \sum_{r=0}^n \frac{r}{^{n}C_r} + \sum_{r=0}^n \frac{n-r}{^{n}C_r} = \sum_{r=0}^n \frac{r + n - r}{^{n}C_r} = n \sum_{r=0}^n \frac{1}{^{n}C_r} \] Since $a_n = \sum_{r=0}^n \frac{1}{^{n}C_r}$, we can substitute it into the equation: \[ 2b_n = n a_n \implies \frac{b_n}{a_n} = \frac{n}{2} \]

Step 4: Conclusion
The ratio of the two series $\frac{b_n}{a_n}$ is mathematically constant at $\frac{n}{2}$. Final Answer: (B)