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

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

$6(n^3 + x^2 + y)$

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

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

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

$= 806$