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) \(\frac{n(n+1)}{2}\)

Answer 2

Let the given expression be denoted by \( S \). We have:

\[ S = \sum_{r=1}^n \frac{r \cdot \binom{n}{r}}{\binom{n}{r-1}} \]

We know that \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \). Therefore,

\[ \frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{\frac{n!}{r!(n-r)!}}{\frac{n!}{(r-1)!(n-r+1)!}} = \frac{(r-1)!(n-r+1)!}{r!(n-r)!} = \frac{n-r+1}{r} \]

Substituting this into the expression for \( S \):

\[ S = \sum_{r=1}^n r \cdot \frac{n-r+1}{r} = \sum_{r=1}^n (n-r+1) = \sum_{r=1}^n (n+1-r) \]

This is an arithmetic series with first term \( a = n \) and last term \( l = 1 \). The number of terms is \( n \).

The sum of an arithmetic series is given by:

\[ S = \frac{n}{2}(a + l) = \frac{n}{2}(n + 1) \]

Therefore,

\[ S = \frac{n(n+1)}{2} \]