Detailed Solution for Binomial Coefficients Ratio Problem

This problem requires a strong understanding of binomial expansion, specifically the general term and the ratio of consecutive binomial coefficients.

---

Key Concepts and Formulas

1. General Term of a Binomial Expansion: For the expansion of $(a+b)^N$, the $(k+1)^{th}$ term, denoted as $T_{k+1}$, is given by: $T_{k+1} = {^N}C_k a^{N-k} b^k$ In our problem, the expansion is $(1+x)^{n+2}$. Here, $a=1$, $b=x$, and $N=n+2$. So, the $(k+1)^{th}$ term is: $T_{k+1} = {^{n+2}}C_k (1)^{n+2-k} (x)^k = {^{n+2}}C_k x^k$ The coefficient of the $(k+1)^{th}$ term is ${^{n+2}}C_k$.

2. Ratio of Consecutive Binomial Coefficients: The ratio of two consecutive binomial coefficients, ${^N}C_k$ and ${^N}C_{k-1}$, is given by: $\frac{{^N}C_k}{{^N}C_{k-1}} = \frac{N-k+1}{k}$ This formula is crucial for solving problems involving ratios of terms in a binomial expansion.

---

Step-by-Step Working

1. Identify the Coefficients of the Consecutive Terms

Let the three consecutive terms be $T_r$, $T_{r+1}$, and $T_{r+2}$. For the expansion of $(1+x)^{n+2}$, the general term $T_{k+1}$ has a coefficient of ${^{n+2}}C_k$. Therefore, the coefficients of $T_r$, $T_{r+1}$, and $T_{r+2}$ are:

• Coefficient of $T_r$: ${^{n+2}}C_{r-1}$ (since $r = (r-1)+1$)
• Coefficient of $T_{r+1}$: ${^{n+2}}C_r$ (since $r+1 = r+1$)
• Coefficient of $T_{r+2}$: ${^{n+2}}C_{r+1}$ (since $r+2 = (r+1)+1$)

We are given that these coefficients are in the ratio $1:3:5$. So, we have: ${^{n+2}}C_{r-1} : {^{n+2}}C_r : {^{n+2}}C_{r+1} = 1:3:5$

2. Formulate Equations using Ratios of Consecutive Coefficients

We can form two separate equations from this given ratio:

• First Ratio: The ratio of the coefficient of $T_{r+1}$ to the coefficient of $T_r$ is $3:1$. $\frac{\text{Coefficient of } T_{r+1}}{\text{Coefficient of } T_r} = \frac{{^{n+2}}C_r}{{^{n+2}}C_{r-1}} = \frac{3}{1} = 3$ Using the ratio formula $\frac{{^N}C_k}{{^N}C_{k-1}} = \frac{N-k+1}{k}$ with $N=n+2$ and $k=r$: $\frac{(n+2)-r+1}{r} = 3$ $\frac{n+3-r}{r} = 3$ Multiplying both sides by $r$: $n+3-r = 3r$ $n - 4r + 3 = 0 \quad \mathbf{(Equation \; 1)}$ Why this step? We use the given ratio to establish a relationship between $n$ and $r$. The ratio formula simplifies the complex binomial coefficients into a linear equation, which is easier to solve.

• Second Ratio: The ratio of the coefficient of $T_{r+2}$ to the coefficient of $T_{r+1}$ is $5:3$. $\frac{\text{Coefficient of } T_{r+2}}{\text{Coefficient of } T_{r+1}} = \frac{{^{n+2}}C_{r+1}}{{^{n+2}}C_r} = \frac{5}{3}$ Using the ratio formula $\frac{{^N}C_k}{{^N}C_{k-1}} = \frac{N-k+1}{k}$ with $N=n+2$ and $k=r+1$: $\frac{(n+2)-(r+1)+1}{r+1} = \frac{5}{3}$ $\frac{n-r+2}{r+1} = \frac{5}{3}$ Cross-multiplying: $3(n-r+2) = 5(r+1)$ $3n - 3r + 6 = 5r + 5$ $3n - 8r + 1 = 0 \quad \mathbf{(Equation \; 2)}$ Why this step? Similar to the first ratio, this gives us another independent linear equation connecting $n$ and $r$. With two linear equations, we can solve for two unknowns.

3. Solve the System of Linear Equations for $n$ and $r$

We have the following system of equations:

1. $n - 4r + 3 = 0$
2. $3n - 8r + 1 = 0$

From Equation 1, we can express $n$ in terms of $r$: $n = 4r - 3$ Substitute this expression for $n$ into Equation 2: $3(4r - 3) - 8r + 1 = 0$ $12r - 9 - 8r + 1 = 0$ $4r - 8 = 0$ $4r = 8$ $r = 2$ Now, substitute the value of $r=2$ back into the expression for $n$: $n = 4(2) - 3$ $n = 8 - 3$ $n = 5$ So, we have found that $n=5$ and $r=2$.

Why this step? Determining the values of $n$ and $r$ is essential because they define the specific binomial expansion and the positions of the consecutive terms whose coefficients we need to sum.

4. Calculate the Sum of the Coefficients

With $n=5$, the binomial expansion is $(1+x)^{n+2} = (1+x)^{5+2} = (1+x)^7$. The three consecutive terms, starting from $T_r$ where $r=2$, are $T_2$, $T_3$, and $T_4$. Their coefficients are:

• Coefficient of $T_2 = {^7}C_{2-1} = {^7}C_1$
• Coefficient of $T_3 = {^7}C_2$
• Coefficient of $T_4 = {^7}C_3$

Now, we calculate the values of these binomial coefficients:

• ${^7}C_1 = \frac{7!}{1!(7-1)!} = \frac{7}{1} = 7$
• ${^7}C_2 = \frac{7!}{2!(7-2)!} = \frac{7 \times 6}{2 \times 1} = 21$
• ${^7}C_3 = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

The sum of these coefficients is: $\text{Sum} = {^7}C_1 + {^7}C_2 + {^7}C_3 = 7 + 21 + 35 = 63$ Why this step? This is the final calculation required by the problem statement. Once $n$ and $r$ are known, we can directly find the specific coefficients and sum them up.

---

Tips and Common Mistakes

• Indexing: Be very careful with the index $k$ in ${^N}C_k$. If a term is the $(k+1)^{th}$ term, its coefficient involves ${^N}C_k$. If you label your terms as $T_r, T_{r+1}, T_{r+2}$, their coefficients will be ${^N}C_{r-1}, {^N}C_r, {^N}C_{r+1}$ respectively. A common error is to use $k$ as the term number instead of the lower index of the combination.
• Ratio Formula Application: Ensure you correctly identify $N$ (the power of the binomial) and $k$ (the lower index of the numerator coefficient) when applying the ratio formula $\frac{{^N}C_k}{{^N}C_{k-1}} = \frac{N-k+1}{k}$.
• Algebraic Errors: Solving the system of linear equations requires careful algebraic manipulation. Double-check your calculations to avoid errors.
• Understanding the Question: The problem asks for the sum of the coefficients, not the terms themselves. The terms would involve $x^{r-1}$, $x^r$, and $x^{r+1}$.

---

Summary and Key Takeaway

This problem effectively tests your understanding of the binomial theorem, particularly how to represent consecutive terms and their coefficients, and how to use the ratio of consecutive binomial coefficients to set up and solve equations. The ability to correctly apply the formula $\frac{{^N}C_k}{{^N}C_{k-1}} = \frac{N-k+1}{k}$ and solve the resulting system of linear equations is paramount. The final answer is the sum of these determined coefficients, which is $63$.

The final answer is $\boxed{\text{63}}$.