if a,b,c and d are the coefficient of four consecutive terms in the expansion of `(1+x)^(n)` then `(a)/(a+b)+(c) /(c+d)=?`

Options

A. `(b)/(b+c)`
B. `(b)/(2(b+c))`
C. `(2b)/(b+c)`
D. `(2c)/(b+c)`

Explanation

To solve the problem, we need to find the value of \(\frac{a}{a+b} + \frac{c}{c+d}\) where \(a\), \(b\), \(c\), and \(d\) are the coefficients of four consecutive terms in the expansion of \((1+x)^n\).

### Step-by-Step Solution:

1. **Identify the Coefficients**:
- The coefficients of the four consecutive terms can be expressed in terms of binomial coefficients:
- Let \(a = \binom{n}{r-1}\)
- Let \(b = \binom{n}{r}\)
- Let \(c = \binom{n}{r+1}\)
- Let \(d = \binom{n}{r+2}\)

2. **Set Up the Expression**:
- We need to evaluate:
\[ \frac{a}{a+b} + \frac{c}{c+d} \]
- Substituting the values of \(a\), \(b\), \(c\), and \(d\):
\[ \frac{\binom{n}{r-1}}{\binom{n}{r-1} + \binom{n}{r}} + \frac{\binom{n}{r+1}}{\binom{n}{r+1} + \binom{n}{r+2}} \]

3. **Simplify the First Term**:
- The first term can be simplified using the property of binomial coefficients:
\[ \frac{\binom{n}{r-1}}{\binom{n}{r-1} + \binom{n}{r}} = \frac{\binom{n}{r-1}}{\binom{n}{r-1} + \binom{n}{r}} = \frac{\binom{n}{r-1}}{\binom{n}{r-1} + \frac{n-r+1}{r}\binom{n}{r-1}} = \frac{1}{1 + \frac{n-r+1}{r}} = \frac{r}{n+1} \]

4. **Simplify the Second Term**:
- Similarly, for the second term:
\[ \frac{\binom{n}{r+1}}{\binom{n}{r+1} + \binom{n}{r+2}} = \frac{\binom{n}{r+1}}{\binom{n}{r+1} + \frac{r+1}{n-r+1}\binom{n}{r+1}} = \frac{1}{1 + \frac{r+1}{n-r+1}} = \frac{n-r+1}{n+1} \]

5. **Combine the Results**:
- Now we can combine the simplified terms:
\[ \frac{r}{n+1} + \frac{n-r+1}{n+1} = \frac{r + n - r + 1}{n+1} = \frac{n + 1}{n + 1} = 1 \]

### Final Answer:
\[ \frac{a}{a+b} + \frac{c}{c+d} = 1 \]