Question

In the expansion of $(1 + x)^n$, $\frac{{}^nC_1}{{}^nC_0} + 2 \cdot \frac{{}^nC_2}{{}^nC_1} + 3 \cdot \frac{{}^nC_3}{{}^nC_2} + \dots + n \cdot \frac{{}^nC_n}{{}^nC_{n-1}}$ is equal to:

• A. $\frac{n(n+1)}{2}$
• B. $\frac{n}{2}$
• C. $\frac{n+1}{2}$
• D. $3n(n+1)$

Answer 1

The Correct Option is A

1. Understand the problem:

We are given the expansion of $(1 + x)^n$ and need to evaluate the sum:

$$\frac{C_1}{C_0} + 2 \frac{C_2}{C_1} + 3 \frac{C_3}{C_2} + \ldots + n \frac{C_n}{C_{n-1}}$$

where $C_k = \binom{n}{k}$, the binomial coefficients.

2. Express the general term:

Each term in the sum is of the form $k \frac{C_k}{C_{k-1}}$. We can simplify this ratio:

$$\frac{C_k}{C_{k-1}} = \frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n - k + 1}{k}$$

This comes from the property of binomial coefficients: $\binom{n}{k} = \frac{n!}{k!(n-k)!}$.

3. Simplify the general term:

Substituting the simplified ratio into the general term:

$$k \frac{C_k}{C_{k-1}} = k \cdot \frac{n - k + 1}{k} = n - k + 1$$

4. Rewrite the sum:

The original sum now becomes:

$$\sum_{k=1}^n (n - k + 1) = \sum_{k=1}^n (n + 1 - k)$$

5. Evaluate the sum:

This is an arithmetic series. Let $m = n + 1$, then the sum is:

$$\sum_{k=1}^n (m - k) = \sum_{k=1}^n m - \sum_{k=1}^n k = n m - \frac{n(n+1)}{2}$$

Substituting back $m = n + 1$:

$$n(n + 1) - \frac{n(n+1)}{2} = \frac{n(n+1)}{2}$$

6. Match the result to the options:

The simplified form $\frac{n(n+1)}{2}$ corresponds to option (A).

Correct Answer: (A) $\frac{n(n+1)}{2}$