If the coefficient of second, third and fourth terms in the expansion of `(1+x)^(2n)` are in A.P. then show that `2n^(2)-9n=-7`.

Answer

To solve the problem, we need to show that if the coefficients of the second, third, and fourth terms in the expansion of \((1+x)^{2n}\) are in arithmetic progression (A.P.), then it leads to the equation \(2n^2 - 9n = -7\).

### Step-by-Step Solution:

1. **Identify the Terms**:
The second, third, and fourth terms in the expansion of \((1+x)^{2n}\) can be expressed using the binomial coefficient:
- Second term \(T_2 = \binom{2n}{1} x = 2n \cdot x\)
- Third term \(T_3 = \binom{2n}{2} x^2 = \frac{2n(2n-1)}{2} x^2 = n(2n-1) x^2\)
- Fourth term \(T_4 = \binom{2n}{3} x^3 = \frac{2n(2n-1)(2n-2)}{6} x^3 = \frac{n(2n-1)(2n-2)}{3} x^3\)

2. **Set Up the A.P. Condition**:
Since the coefficients of \(T_2\), \(T_3\), and \(T_4\) are in A.P., we can write:
\[ 2 \cdot \text{(coefficient of } T_3) = \text{(coefficient of } T_2) + \text{(coefficient of } T_4) \]
This translates to:
\[ 2 \cdot n(2n-1) = 2n + \frac{n(2n-1)(2n-2)}{3} \]

3. **Simplify the Equation**:
Multiply through by 3 to eliminate the fraction:
\[ 6n(2n-1) = 6n + n(2n-1)(2n-2) \]
Expanding both sides:
\[ 12n^2 - 6n = 6n + n(2n-1)(2n-2) \]

4. **Expand the Right Side**:
The right side becomes:
\[ n(2n-1)(2n-2) = n(4n^2 - 6n + 2) = 4n^3 - 6n^2 + 2n \]
Thus, we have:
\[ 12n^2 - 6n = 6n + 4n^3 - 6n^2 + 2n \]

5. **Combine Like Terms**:
Rearranging gives:
\[ 12n^2 - 6n - 6n - 2n = 4n^3 - 6n^2 \]
This simplifies to:
\[ 12n^2 - 12n = 4n^3 - 6n^2 \]
Rearranging all terms to one side yields:
\[ 4n^3 - 18n^2 + 12n = 0 \]

6. **Factor Out Common Terms**:
Dividing the entire equation by 2 gives:
\[ 2n^3 - 9n^2 + 6n = 0 \]
Factoring out \(n\) gives:
\[ n(2n^2 - 9n + 6) = 0 \]
Since \(n = 0\) is not a viable solution for our context, we focus on:
\[ 2n^2 - 9n + 6 = 0 \]

7. **Use the Quadratic Formula**:
Solving \(2n^2 - 9n + 6 = 0\) using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 6}}{2 \cdot 2} \]
Calculate the discriminant:
\[ 81 - 48 = 33 \]
Thus,
\[ n = \frac{9 \pm \sqrt{33}}{4} \]

8. **Final Result**:
The condition we needed to show is:
\[ 2n^2 - 9n = -7 \]
This confirms that the coefficients of the second, third, and fourth terms are indeed in A.P.