Key Concepts and Formula

The core concept for this problem is the Binomial Theorem, which describes the algebraic expansion of powers of a binomial. For any non-negative integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{r=0}^n {}^n C_r a^{n-r} b^r$ The general term, or the $(r+1)^{th}$ term, in the expansion of $(a+b)^n$ is given by: $T_{r+1} = {}^n C_r a^{n-r} b^r$ In this specific problem, we are expanding $(x+y)^n$. Therefore, we use $a=x$ and $b=y$, making the general term: $T_{r+1} = {}^n C_r x^{n-r} y^r$

Problem Setup: Expressing Given Terms

We are given the values of the second, third, and fourth terms of the expansion $(x+y)^n$. We will use the general term formula to write these terms in terms of $n$, $x$, and $y$.

1. Second Term ($T_2$): For $T_2$, we have $r+1 = 2$, so $r=1$. $T_2 = {}^n C_1 x^{n-1} y^1 = 135 \quad \text{(Equation 1)}$

2. Third Term ($T_3$): For $T_3$, we have $r+1 = 3$, so $r=2$. $T_3 = {}^n C_2 x^{n-2} y^2 = 30 \quad \text{(Equation 2)}$

3. Fourth Term ($T_4$): For $T_4$, we have $r+1 = 4$, so $r=3$. $T_4 = {}^n C_3 x^{n-3} y^3 = \frac{10}{3} \quad \text{(Equation 3)}$

Step 1: Forming Ratios of Consecutive Terms to Find $n$

A highly effective strategy when dealing with consecutive terms in a binomial expansion is to take their ratios. This simplifies the expressions significantly, allowing us to find relationships between $n$, $x$, and $y$.

Ratio of $T_2$ to $T_3$: We divide Equation 1 by Equation 2: $\frac{T_2}{T_3} = \frac{{}^n C_1 x^{n-1} y^1}{{}^n C_2 x^{n-2} y^2} = \frac{135}{30}$ First, simplify the numerical ratio: $\frac{135}{30} = \frac{27 \times 5}{6 \times 5} = \frac{27}{6} = \frac{9}{2}$.

Now, simplify the terms involving $n$, $x$, and $y$. Recall the property of binomial coefficients: $\frac{{}^n C_r}{{}^n C_{r+1}} = \frac{r+1}{n-r}$. Here, we have $\frac{{}^n C_1}{{}^n C_2}$, which corresponds to $r=1$. So, $\frac{{}^n C_1}{{}^n C_2} = \frac{1+1}{n-1} = \frac{2}{n-1}$. Also, $\frac{x^{n-1}}{x^{n-2}} = x^{n-1-(n-2)} = x^1 = x$, and $\frac{y^1}{y^2} = y^{1-2} = y^{-1} = \frac{1}{y}$.

Combining these, the ratio becomes: $\frac{2}{n-1} \cdot \frac{x}{y} = \frac{9}{2} \quad \text{(Equation A)}$

Ratio of $T_3$ to $T_4$: Next, we divide Equation 2 by Equation 3: $\frac{T_3}{T_4} = \frac{{}^n C_2 x^{n-2} y^2}{{}^n C_3 x^{n-3} y^3} = \frac{30}{\frac{10}{3}}$ Simplify the numerical ratio: $\frac{30}{\frac{10}{3}} = 30 \times \frac{3}{10} = 3 \times 3 = 9$.

Now, simplify the terms involving $n$, $x$, and $y$. For $\frac{{}^n C_2}{{}^n C_3}$, we use the same property with $r=2$: $\frac{{}^n C_2}{{}^n C_3} = \frac{2+1}{n-2} = \frac{3}{n-2}$. Also, $\frac{x^{n-2}}{x^{n-3}} = x^{n-2-(n-3)} = x^1 = x$, and $\frac{y^2}{y^3} = y^{2-3} = y^{-1} = \frac{1}{y}$.

Combining these, the ratio becomes: $\frac{3}{n-2} \cdot \frac{x}{y} = 9 \quad \text{(Equation B)}$

Solving for $n$: Now we have a system of two equations (A and B) with two unknowns, $n$ and $\frac{x}{y}$. We can solve this system.

From Equation A, isolate $\frac{x}{y}$: $\frac{x}{y} = \frac{9}{2} \cdot \frac{n-1}{2} = \frac{9(n-1)}{4}$

From Equation B, isolate $\frac{x}{y}$: $\frac{x}{y} = 9 \cdot \frac{n-2}{3} = 3(n-2)$

Now, equate the two expressions for $\frac{x}{y}$: $\frac{9(n-1)}{4} = 3(n-2)$ To eliminate the fraction, multiply both sides by 4: $9(n-1) = 12(n-2)$ Distribute the terms: $9n - 9 = 12n - 24$ Rearrange to solve for $n$: $24 - 9 = 12n - 9n$ $15 = 3n$ $n = \frac{15}{3} = 5$ Thus, the value of $n$ is 5.

Step 2: Finding $x$ and $y$

With $n=5$, we can now find the values of $x$ and $y$.

First, substitute $n=5$ back into either Equation A or Equation B to find the ratio $\frac{x}{y}$. Using Equation B is simpler: $\frac{3}{n-2} \cdot \frac{x}{y} = 9$ $\frac{3}{5-2} \cdot \frac{x}{y} = 9$ $\frac{3}{3} \cdot \frac{x}{y} = 9$ $1 \cdot \frac{x}{y} = 9$ $\frac{x}{y} = 9$ This implies $x = 9y$.

Now, substitute $n=5$ and $x=9y$ into one of the original term equations (Equation 1, 2, or 3) to find the explicit values of $x$ and $y$. Let's use Equation 1 ($T_2$): $T_2 = {}^n C_1 x^{n-1} y^1 = 135$ Substitute $n=5$: ${}^5 C_1 x^{5-1} y^1 = 135$ We know ${}^5 C_1 = 5$: $5 x^4 y = 135$ Divide by 5: $x^4 y = 27$ Now substitute $x=9y$: $(9y)^4 y = 27$ $9^4 y^4 y = 27$ $9^4 y^5 = 27$ Express 9 and 27 as powers of 3 ($9=3^2$, $27=3^3$): $(3^2)^4 y^5 = 3^3$ $3^8 y^5 = 3^3$ Isolate $y^5$: $y^5 = \frac{3^3}{3^8} = 3^{3-8} = 3^{-5}$ $y^5 = \left(\frac{1}{3}\right)^5$ Therefore, $y = \frac{1}{3}$.

Finally, use $x=9y$ to find $x$: $x = 9 \cdot \frac{1}{3} = 3$

So, we have determined the values: $n=5$, $x=3$, and $y=\frac{1}{3}$.

Tip for avoiding common mistakes: When solving for $x$ and $y$, always substitute back into one of the original term equations ($T_2$, $T_3$, or $T_4$) rather than the derived ratio equations. This minimizes the chance of propagating errors from earlier algebraic manipulations. Also, be meticulous with exponent rules and arithmetic, especially when dealing with fractions.

Step 3: Calculating the Final Expression

The problem asks for the value of the expression $6(n^3+x^2+y)$. Substitute the values we found: $n=5$, $x=3$, and $y=\frac{1}{3}$.

$6(n^3+x^2+y) = 6\left( (5)^3 + (3)^2 + \frac{1}{3} \right)$ Calculate the powers: $= 6\left( 125 + 9 + \frac{1}{3} \right)$ Add the integer terms: $= 6\left( 134 + \frac{1}{3} \right)$ Now, distribute the 6 to both terms inside the parenthesis: $= (6 \times 134) + \left( 6 \times \frac{1}{3} \right)$ $= 804 + 2$ $= 806$

Summary and Key Takeaway

This problem is an excellent illustration of how to leverage the properties of binomial expansions, particularly the ratios of consecutive terms. By systematically setting up equations from the given information and wisely taking ratios, we were able to simplify the problem into manageable steps for solving $n$, $x$, and $y$. The key takeaway is that when faced with problems involving multiple consecutive terms of a binomial expansion, consider using the ratio of terms (e.g., $\frac{T_{r+1}}{T_r}$) to simplify the system of equations. Always double-check calculations and substitutions to ensure accuracy.