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

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

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

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

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

\[ = 806 \]