If `a_r` is the coefficient of `x^r` in the expansion of `(1+x)^n` then `a_1/a_0 + 2.a_2/a_1 + 3.a_3/a_2 + …..+n.(a_n)/(a_(n-1))` =

Options

A. `(n(n+1))/(2)`
B. `(n(n+3))/(2)`
C. `(n(n-1))/(2)`
D. `n^2`

Answer

To solve the problem, we need to find the value of the expression:

\[ \frac{a_1}{a_0} + 2 \cdot \frac{a_2}{a_1} + 3 \cdot \frac{a_3}{a_2} + \ldots + n \cdot \frac{a_n}{a_{n-1}} \]

where \( a_r \) is the coefficient of \( x^r \) in the expansion of \( (1+x)^n \).

### Step 1: Identify the coefficients \( a_r \)

The coefficients \( a_r \) can be expressed using the binomial coefficient:

\[ a_r = \binom{n}{r} \]

Thus, we have:

- \( a_0 = \binom{n}{0} = 1 \)
- \( a_1 = \binom{n}{1} = n \)
- \( a_2 = \binom{n}{2} = \frac{n(n-1)}{2} \)
- \( a_3 = \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \)
- ...
- \( a_n = \binom{n}{n} = 1 \)

### Step 2: Calculate \( \frac{a_r}{a_{r-1}} \)

Next, we calculate \( \frac{a_r}{a_{r-1}} \):

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

### Step 3: Substitute into the expression

Now substituting \( \frac{a_r}{a_{r-1}} \) into the original expression:

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

### Step 4: Simplify the summation

The sum can be simplified:

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

Calculating each part:

1. The first part: \( \sum_{r=1}^{n} (n+1) = n(n+1) \)
2. The second part: \( \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \)

Putting it together:

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

### Final Result

Thus, the required sum is:

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