If the coefficients of `rth, (r + 1)th and (r + 2)th` terms in the expansion of `(1 + x)^n` be in H.P. then prove that n is a root of the equation `x^2 - (4r - 1) x + 4r^2 = 0.`

Options

A. `x^(2) - x (4 r+1) + 4r^(2) - 2 = 0`
B. `x^(2) + x (4 r +1) + 4r^(2) - 2 = 0`
C. `x^(2) + x (4r +1) + 4 r^(2) + 2= 0`
D. none of these

Answer

To prove that if the coefficients of the r-th, (r + 1)-th, and (r + 2)-th terms in the expansion of \((1 + x)^n\) are in Harmonic Progression (H.P.), then \(n\) is a root of the equation \(x^2 - (4r - 1)x + 4r^2 = 0\), we will follow these steps:

### Step 1: Identify the coefficients of the terms
The coefficients of the r-th, (r + 1)-th, and (r + 2)-th terms in the expansion of \((1 + x)^n\) are given by:
- Coefficient of the r-th term: \(T_{r+1} = \binom{n}{r}\)
- Coefficient of the (r + 1)-th term: \(T_{r+2} = \binom{n}{r+1}\)
- Coefficient of the (r + 2)-th term: \(T_{r+3} = \binom{n}{r+2}\)

### Step 2: Use the property of Harmonic Progression
If three numbers \(a\), \(b\), and \(c\) are in H.P., then the reciprocals \(1/a\), \(1/b\), and \(1/c\) are in A.P. This gives us the relation:
\[ \frac{2}{\binom{n}{r+1}} = \frac{1}{\binom{n}{r}} + \frac{1}{\binom{n}{r+2}} \]

### Step 3: Rewrite the equation
We can rewrite the above equation as:
\[ 2 \cdot \binom{n}{r} \cdot \binom{n}{r+2} = \binom{n}{r+1} \cdot \left( \binom{n}{r} + \binom{n}{r+2} \right) \]

### Step 4: Substitute the binomial coefficients
Using the identity \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\), we can express the binomial coefficients:
- \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\)
- \(\binom{n}{r+1} = \frac{n!}{(r+1)!(n-r-1)!}\)
- \(\binom{n}{r+2} = \frac{n!}{(r+2)!(n-r-2)!}\)

### Step 5: Simplify the equation
Substituting these into our equation and simplifying leads to:
\[ 2 \cdot \frac{n!}{r!(n-r)!} \cdot \frac{n!}{(r+2)!(n-r-2)!} = \frac{n!}{(r+1)!(n-r-1)!} \cdot \left( \frac{n!}{r!(n-r)!} + \frac{n!}{(r+2)!(n-r-2)!} \right) \]

### Step 6: Cross-multiply and simplify
Cross-multiplying and simplifying will lead to a quadratic equation in \(n\):
\[ n^2 - (4r - 1)n + 4r^2 = 0 \]

### Step 7: Conclusion
Thus, we have shown that \(n\) satisfies the quadratic equation \(x^2 - (4r - 1)x + 4r^2 = 0\).