Key Concept: Ratio of Consecutive Binomial Coefficients

In the binomial expansion of $(x+y)^n$, the general term is given by $T_{k+1} = ^n C_k x^{n-k} y^k$. For the expansion of $(x+1)^n$, the general term is $T_{k+1} = ^n C_k x^k (1)^{n-k} = ^n C_k x^k$. Therefore, the coefficient of $x^k$ is $^n C_k$.

A crucial property for solving problems involving consecutive binomial coefficients is their ratio. The ratio of two consecutive binomial coefficients, $^n C_k$ and $^n C_{k-1}$, is given by the formula: $\frac{^n C_k}{^n C_{k-1}} = \frac{n-k+1}{k}$ From this, we can also derive the following useful forms:

• $\frac{^n C_{k-1}}{^n C_k} = \frac{k}{n-k+1}$
• $\frac{^n C_k}{^n C_{k+1}} = \frac{k+1}{n-k}$

Step-by-Step Solution

1. Identify the Consecutive Coefficients and their Ratios

Let the three consecutive coefficients in the binomial expansion of $(x+1)^n$ be $C_{r-1}$, $C_r$, and $C_{r+1}$. These correspond to the coefficients of $x^{r-1}$, $x^r$, and $x^{r+1}$ respectively. Thus, these coefficients are $^n C_{r-1}$, $^n C_r$, and $^n C_{r+1}$.

The problem states that these coefficients are in the ratio $2:15:70$: $^n C_{r-1} : ^n C_r : ^n C_{r+1} = 2 : 15 : 70$ From this, we can extract two individual ratio equations: a) $\frac{^n C_{r-1}}{^n C_r} = \frac{2}{15}$ b) $\frac{^n C_r}{^n C_{r+1}} = \frac{15}{70}$

2. Formulate Equations using the Ratio Property

• Using the first ratio (a): We have $\frac{^n C_{r-1}}{^n C_r} = \frac{2}{15}$. To simplify this ratio, we use the formula $\frac{^n C_{k-1}}{^n C_k} = \frac{k}{n-k+1}$. Here, our index for $k$ is $r$. Substituting $k=r$: $\frac{r}{n-r+1} = \frac{2}{15}$ Explanation: This step translates the combinatorial ratio into an algebraic equation involving $n$ and $r$, which are the unknowns we need to find. Now, cross-multiply to eliminate the denominators: $15r = 2(n-r+1)$ $15r = 2n - 2r + 2$ Rearrange the terms to get a linear equation: $17r = 2n + 2 \quad \text{... (Equation 1)}$

• Using the second ratio (b): We have $\frac{^n C_r}{^n C_{r+1}} = \frac{15}{70}$. This ratio can be simplified by dividing both numerator and denominator by 5, giving $\frac{3}{14}$. So, $\frac{^n C_r}{^n C_{r+1}} = \frac{3}{14}$. Now, we use the formula $\frac{^n C_k}{^n C_{k+1}} = \frac{k+1}{n-k}$. Here, our index for $k$ is $r$. Substituting $k=r$: $\frac{r+1}{n-r} = \frac{3}{14}$ Explanation: Similar to the first ratio, this step converts the second combinatorial ratio into another algebraic equation in terms of $n$ and $r$. Cross-multiply: $14(r+1) = 3(n-r)$ $14r + 14 = 3n - 3r$ Rearrange the terms: $17r = 3n - 14 \quad \text{... (Equation 2)}$

3. Solve for n and r

We now have a system of two linear equations:

1. $17r = 2n + 2$
2. $17r = 3n - 14$

• Since both Equation 1 and Equation 2 are equal to $17r$, we can equate their right-hand sides to solve for $n$: $2n + 2 = 3n - 14$ Explanation: By setting the expressions for $17r$ equal to each other, we eliminate the variable $r$ and obtain a single equation with only $n$, which is straightforward to solve. Subtract $2n$ from both sides: $2 = n - 14$ Add 14 to both sides: $n = 16$

• Now that we have the value of $n$, substitute $n=16$ into either Equation 1 or Equation 2 to find $r$. Let's use Equation 1: $17r = 2(16) + 2$ $17r = 32 + 2$ $17r = 34$ Divide by 17: $r = \frac{34}{17}$ $r = 2$

• Verification: For binomial coefficients $^n C_k$, we must have $n \ge k \ge 0$. Here, our coefficients are $^n C_{r-1}$, $^n C_r$, $^n C_{r+1}$. With $n=16$ and $r=2$, the indices are $1, 2, 3$. All these indices are valid: $0 \le 1 < 2 < 3 \le 16$. This confirms our values for $n$ and $r$ are consistent.

4. Identify and Calculate the Specific Coefficients

With $n=16$ and $r=2$, the three consecutive coefficients are $^n C_{r-1}$, $^n C_r$, and $^n C_{r+1}$. Substituting the values of $n$ and $r$, these are:

• $^16 C_{2-1} = ^16 C_1$
• $^16 C_2$
• $^16 C_{2+1} = ^16 C_3$

Now, let's calculate the values of these binomial coefficients:

• $^16 C_1 = \frac{16!}{1!(16-1)!} = \frac{16}{1} = 16$
• $^16 C_2 = \frac{16!}{2!(16-2)!} = \frac{16 \times 15}{2 \times 1} = 8 \times 15 = 120$
• $^16 C_3 = \frac{16!}{3!(16-3)!} = \frac{16 \times 15 \times 14}{3 \times 2 \times 1} = 8 \times 5 \times 14 = 560$

5. Calculate the Average of these Coefficients

The problem asks for the average of these three coefficients. The average is the sum of the coefficients divided by the count (which is 3): $\text{Average} = \frac{^16 C_1 + ^16 C_2 + ^16 C_3}{3}$ $\text{Average} = \frac{16 + 120 + 560}{3}$ $\text{Average} = \frac{696}{3}$ $\text{Average} = 232$

Tips for Success & Common Mistakes

• Correct Formula Application: Ensure you use the precise ratio formula for the given consecutive terms. A common mistake is to invert the formula or use the wrong index.
• Algebraic Precision: Solving the system of linear equations for $n$ and $r$ requires careful algebraic manipulation. Double-check all calculations to avoid errors.
• Validity of $n$ and $r$: After finding $n$ and $r$, always perform a quick check to ensure they are non-negative integers and that $r \le n$. This confirms the coefficients are valid.
• Understanding "Consecutive": In binomial expansion, "consecutive coefficients" usually implies $^n C_k$, $^n C_{k+1}$, $^n C_{k+2}$ or $^n C_{k-1}$, $^n C_k$, $^n C_{k+1}$ as used here.

Summary & Key Takeaway

This problem effectively tests the understanding and application of the ratio property of binomial coefficients. The methodical approach involves:

1. Translating the given ratios of coefficients into algebraic equations using the specific formulas.
2. Solving the resulting system of linear equations to find the values of $n$ (the power of the binomial) and $r$ (the index of the first of the three consecutive terms).
3. Calculating the actual values of the identified coefficients.
4. Finally, computing their average.

The ability to accurately apply the combinatorial identities and perform algebraic manipulations is crucial for solving such problems efficiently. The average of the three consecutive coefficients is $232$.