Question

If the second, third, and fourth terms in the expansion of \( (x + y)^n \) are \( 135 \), \( 30 \), and \( \frac{10}{3} \), respectively, then \( 6\left(n^3 + x^2 + y\right) \) is equal to ______.

Answer 1

Correct Answer: 806

The given terms are:

\(nC_1 x^{n-1} y = 135\)       (i)

\(nC_2 x^{n-2} y^2 = 30\)        (ii)

\(nC_3 x^{n-3} y^3 = \frac{10}{3}\)        (iii)

Step 1: Using (i) and (ii)

\(\frac{nC_1 x}{nC_2 y} = \frac{9}{2}\)      (iv)

Step 2: Using (ii) and (iii)

\(\frac{nC_2 x}{nC_3 y} = 9\)     (v)

From (iv) and (v), solve for \(nC_2\):

\[ \frac{nC_1 \cdot nC_3}{(nC_2)^2} = \frac{1}{2} \]

Substitute \(nC_1 = n\), \(nC_2 = \frac{n(n-1)}{2}\), \(nC_3 = \frac{n(n-1)(n-2)}{6}\):

\[ \frac{n \cdot \frac{n(n-1)(n-2)}{6}}{\left(\frac{n(n-1)}{2}\right)^2} = \frac{1}{2} \]

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

\[ 2n(n-2) = 3(n-1) \]

\[ 2n^2 - 7n + 3 = 0 \]

\[ n = 5 \quad \text{(as \(n\) is a positive integer)} \]

Step 3: Solving for \(x\) and \(y\)

From (v):

\[ \frac{x}{y} = 9 \implies x = 9y \]

Substitute in (i):

\[ 5C_1 (9y)^4 y = 135 \]

\[ 5C_1 \cdot 94 y^5 = 135 \]

\[ 5 \cdot 81 \cdot 9 \cdot y^5 = 135 \]

\[ y = \frac{1}{3}, \quad x = 3 \]

Step 4: Calculating the final expression

\[ 6(n^3 + x^2 + y) \]

\[ = 6(5^3 + 3^2 + \frac{1}{3}) \]

\[ = 6(125 + 9 + \frac{1}{3}) \]

\[ = 6(134 + \frac{1}{3}) \]

\[ = 806 \]

Answer 2

We are given that the second, third, and fourth terms in the binomial expansion of \( (x + y)^n \) are \( 135 \), \( 30 \), and \( \frac{10}{3} \), respectively. We need to find the value of the expression \( 6\left(n^3 + x^2 + y\right) \).

Concept Used:

The general term (the \( (r+1) \)-th term) in the binomial expansion of \( (x + y)^n \) is given by the formula:

\[ T_{r+1} = \binom{n}{r} x^{n-r} y^r \]

where \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \).

To solve for the unknowns \( n, x, y \), we can set up a system of equations for the given terms. A useful technique is to take the ratio of consecutive terms, which simplifies the expressions by canceling common factors.

\[ \frac{T_{r+1}}{T_r} = \frac{\binom{n}{r} x^{n-r} y^r}{\binom{n}{r-1} x^{n-r+1} y^{r-1}} = \frac{n-r+1}{r} \frac{y}{x} \]

Step-by-Step Solution:

Step 1: Write the equations for the 2nd, 3rd, and 4th terms.

The second term (\( r=1 \)) is \( T_2 \):

\[ T_2 = \binom{n}{1} x^{n-1} y^1 = n x^{n-1} y = 135 \quad \cdots(1) \]

The third term (\( r=2 \)) is \( T_3 \):

\[ T_3 = \binom{n}{2} x^{n-2} y^2 = \frac{n(n-1)}{2} x^{n-2} y^2 = 30 \quad \cdots(2) \]

The fourth term (\( r=3 \)) is \( T_4 \):

\[ T_4 = \binom{n}{3} x^{n-3} y^3 = \frac{n(n-1)(n-2)}{6} x^{n-3} y^3 = \frac{10}{3} \quad \cdots(3) \]

Step 2: Divide equation (2) by equation (1) to find a relation between \( n, x, y \).

\[ \frac{T_3}{T_2} = \frac{\frac{n(n-1)}{2} x^{n-2} y^2}{n x^{n-1} y} = \frac{30}{135} \] \[ \frac{n-1}{2} \cdot \frac{y}{x} = \frac{2}{9} \implies \frac{y}{x} = \frac{4}{9(n-1)} \quad \cdots(A) \]

Step 3: Divide equation (3) by equation (2).

\[ \frac{T_4}{T_3} = \frac{\frac{n(n-1)(n-2)}{6} x^{n-3} y^3}{\frac{n(n-1)}{2} x^{n-2} y^2} = \frac{10/3}{30} \] \[ \frac{n-2}{3} \cdot \frac{y}{x} = \frac{1}{9} \implies \frac{y}{x} = \frac{3}{9(n-2)} = \frac{1}{3(n-2)} \quad \cdots(B) \]

Step 4: Equate the expressions for \( \frac{y}{x} \) from equations (A) and (B) to find \( n \).

\[ \frac{4}{9(n-1)} = \frac{1}{3(n-2)} \] \[ 4 \cdot 3(n-2) = 9(n-1) \] \[ 12(n-2) = 9(n-1) \implies 4(n-2) = 3(n-1) \] \[ 4n - 8 = 3n - 3 \implies n = 5 \]

Step 5: Substitute \( n=5 \) into equation (B) to find the ratio \( \frac{y}{x} \).

\[ \frac{y}{x} = \frac{1}{3(5-2)} = \frac{1}{3(3)} = \frac{1}{9} \implies y = \frac{x}{9} \]

Step 6: Substitute \( n=5 \) and \( y = \frac{x}{9} \) into equation (1) to find \( x \).

\[ n x^{n-1} y = 135 \] \[ 5 \cdot x^{5-1} \cdot \left(\frac{x}{9}\right) = 135 \] \[ \frac{5}{9} x^5 = 135 \] \[ x^5 = \frac{135 \times 9}{5} = 27 \times 9 = 243 \] \[ x^5 = 3^5 \implies x = 3 \]

Now, we find \( y \):

\[ y = \frac{x}{9} = \frac{3}{9} = \frac{1}{3} \]

Final Computation & Result:

Step 7: We have found \( n = 5 \), \( x = 3 \), and \( y = \frac{1}{3} \). Now, we compute the value of the expression \( 6\left(n^3 + x^2 + y\right) \).

\[ n^3 = 5^3 = 125 \] \[ x^2 = 3^2 = 9 \] \[ y = \frac{1}{3} \]

Step 8: Substitute these values into the expression.

\[ 6\left(n^3 + x^2 + y\right) = 6\left(125 + 9 + \frac{1}{3}\right) \] \[ = 6\left(134 + \frac{1}{3}\right) \] \[ = 6\left(\frac{134 \times 3 + 1}{3}\right) = 6\left(\frac{402 + 1}{3}\right) = 6\left(\frac{403}{3}\right) \] \[ = 2 \times 403 = 806 \]

Therefore, the value of \( 6\left(n^3 + x^2 + y\right) \) is 806.