The 2nd, 3rd and 4th terms in the expansion of `(x+y)^(n)` are `240, 720 and 1080` respectively, find the values of `x,y and n`.

Answer

To solve the problem, we need to find the values of \( x \), \( y \), and \( n \) given that the 2nd, 3rd, and 4th terms in the expansion of \( (x+y)^n \) are 240, 720, and 1080 respectively.

### Step 1: Write the expressions for the terms
The \( r \)-th term in the expansion of \( (x+y)^n \) is given by:
\[ T_r = \binom{n}{r-1} y^{r-1} x^{n-r+1} \]
Thus, we can write:
- The 2nd term \( T_2 = \binom{n}{1} y^1 x^{n-1} = n y x^{n-1} = 240 \)
- The 3rd term \( T_3 = \binom{n}{2} y^2 x^{n-2} = \frac{n(n-1)}{2} y^2 x^{n-2} = 720 \)
- The 4th term \( T_4 = \binom{n}{3} y^3 x^{n-3} = \frac{n(n-1)(n-2)}{6} y^3 x^{n-3} = 1080 \)

### Step 2: Set up the equations
From the terms we have:
1. \( n y x^{n-1} = 240 \) (Equation 1)
2. \( \frac{n(n-1)}{2} y^2 x^{n-2} = 720 \) (Equation 2)
3. \( \frac{n(n-1)(n-2)}{6} y^3 x^{n-3} = 1080 \) (Equation 3)

### Step 3: Simplify the equations
From Equation 1:
\[ n y x^{n-1} = 240 \implies y = \frac{240}{n x^{n-1}} \tag{4} \]

Substituting \( y \) from Equation 4 into Equation 2:
\[ \frac{n(n-1)}{2} \left(\frac{240}{n x^{n-1}}\right)^2 x^{n-2} = 720 \]
Simplifying gives:
\[ \frac{n(n-1)}{2} \cdot \frac{57600}{n^2 x^{2(n-1)}} x^{n-2} = 720 \]
\[ \frac{n-1}{2n} \cdot \frac{57600}{x^{n}} = 720 \]
\[ \frac{n-1}{n} \cdot 57600 = 1440 x^{n} \]
\[ n-1 = \frac{1440 n x^{n}}{57600} = \frac{n x^{n}}{40} \tag{5} \]

Now, substituting \( y \) from Equation 4 into Equation 3:
\[ \frac{n(n-1)(n-2)}{6} \left(\frac{240}{n x^{n-1}}\right)^3 x^{n-3} = 1080 \]
This simplifies to:
\[ \frac{n(n-1)(n-2)}{6} \cdot \frac{13824000}{n^3 x^{3(n-1)}} x^{n-3} = 1080 \]
\[ \frac{n(n-1)(n-2)}{6} \cdot \frac{13824000}{n^3 x^{2n-3}} = 1080 \]
\[ \frac{(n-1)(n-2)}{n^2} \cdot \frac{13824000}{6} = 1080 x^{2n-3} \]
\[ (n-1)(n-2) = \frac{1080 \cdot 6 n^2 x^{2n-3}}{13824000} \tag{6} \]

### Step 4: Solve the equations
Now we have two equations (5) and (6) that we can solve simultaneously.

From Equation 5, we can express \( x^{n} \) in terms of \( n \):
\[ x^{n} = \frac{40(n-1)}{n} \]

Substituting \( x^{n} \) back into Equation 6 will give us a polynomial in terms of \( n \).

After solving these equations, we find:
- \( n = 5 \)
- \( x = 2 \)
- \( y = 3 \)

### Step 5: Verify the solution
Substituting \( n = 5 \), \( x = 2 \), and \( y = 3 \) back into the original equations:
1. \( 5 \cdot 3 \cdot 2^{4} = 5 \cdot 3 \cdot 16 = 240 \) (Correct)
2. \( \frac{5 \cdot 4}{2} \cdot 3^{2} \cdot 2^{3} = 10 \cdot 9 \cdot 8 = 720 \) (Correct)
3. \( \frac{5 \cdot 4 \cdot 3}{6} \cdot 3^{3} \cdot 2^{2} = 10 \cdot 27 \cdot 4 = 1080 \) (Correct)

Thus, the values of \( x \), \( y \), and \( n \) are:
\[ \boxed{x = 2, y = 3, n = 5} \]