If a. b, c and d are the coefficients of 2nd, 3rd, 4th and 5th terms respectively in the binomial expansion of `(1+x)^n`, then prove that `a/(a+b) + c/(c+d) = 2b/(b+c)`

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 coefficients of the 2nd, 3rd, 4th, and 5th terms in the binomial expansion of \((1+x)^n\) and prove that

\[ \frac{a}{a+b} + \frac{c}{c+d} = \frac{2b}{b+c} \]

where \(a\), \(b\), \(c\), and \(d\) are the coefficients of the 2nd, 3rd, 4th, and 5th terms respectively.

### Step 1: Identify the coefficients

The coefficients of the terms in the binomial expansion \((1+x)^n\) can be expressed using the binomial coefficient \( \binom{n}{r} \):

- The 2nd term (coefficient \(a\)):
\[ a = \binom{n}{1} = n \]

- The 3rd term (coefficient \(b\)):
\[ b = \binom{n}{2} = \frac{n(n-1)}{2} \]

- The 4th term (coefficient \(c\)):
\[ c = \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]

- The 5th term (coefficient \(d\)):
\[ d = \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{24} \]

### Step 2: Substitute coefficients into the equation

Now we need to substitute these coefficients into the left-hand side of the equation we want to prove:

\[ \frac{a}{a+b} + \frac{c}{c+d} \]

Substituting \(a\), \(b\), \(c\), and \(d\):

\[ \frac{n}{n + \frac{n(n-1)}{2}} + \frac{\frac{n(n-1)(n-2)}{6}}{\frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)}{24}} \]

### Step 3: Simplify the first term

For the first term:

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

### Step 4: Simplify the second term

For the second term:

\[ \frac{\frac{n(n-1)(n-2)}{6}}{\frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)}{24}} \]

Finding a common denominator for the terms in the denominator:

\[ \frac{n(n-1)(n-2)}{6} + \frac{n(n-1)(n-2)(n-3)}{24} = \frac{4n(n-1)(n-2) + n(n-1)(n-2)(n-3)}{24} \]

This simplifies to:

\[ \frac{n(n-1)(n-2)(4 + (n-3))}{24} = \frac{n(n-1)(n-2)(n+1)}{24} \]

Thus, the second term simplifies to:

\[ \frac{n(n-1)(n-2)}{6} \cdot \frac{24}{n(n-1)(n-2)(n+1)} = \frac{4}{n+1} \]

### Step 5: Combine the two terms

Now we combine both simplified terms:

\[ \frac{2n}{n^2 + n} + \frac{4}{n+1} \]

Finding a common denominator:

\[ \frac{2n(n+1) + 4(n^2 + n)}{(n^2 + n)(n+1)} \]

This simplifies to:

\[ \frac{2n^2 + 2n + 4n^2 + 4n}{(n^2 + n)(n+1)} = \frac{6n^2 + 6n}{(n^2 + n)(n+1)} = \frac{6n(n+1)}{(n^2 + n)(n+1)} = \frac{6n}{n^2 + n} \]

### Step 6: Prove the right-hand side

Now we need to show that:

\[ \frac{6n}{n^2 + n} = \frac{2b}{b+c} \]

Substituting \(b\) and \(c\):

\[ b = \frac{n(n-1)}{2}, \quad c = \frac{n(n-1)(n-2)}{6} \]

Calculating \(b+c\):

\[ b+c = \frac{n(n-1)}{2} + \frac{n(n-1)(n-2)}{6} \]

Finding a common denominator gives:

\[ \frac{3n(n-1) + n(n-1)(n-2)}{6} = \frac{n(n-1)(3 + (n-2))}{6} = \frac{n(n-1)(n+1)}{6} \]

Thus,

\[ \frac{2b}{b+c} = \frac{2 \cdot \frac{n(n-1)}{2}}{\frac{n(n-1)(n+1)}{6}} = \frac{n(n-1) \cdot 6}{n(n-1)(n+1)} = \frac{6}{n+1} \]

### Conclusion

Both sides are equal, proving the required identity:

\[ \frac{a}{a+b} + \frac{c}{c+d} = \frac{2b}{b+c} \]