Key Concept: Binomial Expansion and Ratios of Coefficients

The binomial theorem states that for any positive integer $n$, the expansion of $(a+b)^n$ is given by: $(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ is the binomial coefficient.

For the given expansion $(1 + 1/x)^n$, we can treat $a=1$ and $b=1/x$. The general term is: $T_{k+1} = \binom{n}{k} (1)^{n-k} \left(\frac{1}{x}\right)^k = \binom{n}{k} x^{-k}$ The problem states the expansion is in "increasing powers of x". To achieve this, we consider the terms as coefficients of $x^p$. If we write $(1+1/x)^n = \left(\frac{x+1}{x}\right)^n = \frac{1}{x^n} (x+1)^n$. Then, $(x+1)^n = \sum_{k=0}^n \binom{n}{k} x^k (1)^{n-k} = \sum_{k=0}^n \binom{n}{k} x^k$. So, $(1+1/x)^n = \sum_{k=0}^n \binom{n}{k} x^{k-n}$. As $k$ increases from $0$ to $n$, the power $k-n$ increases from $-n$ to $0$. The coefficients of these terms are $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$. Let three consecutive coefficients in this expansion be $\binom{n}{r-1}$, $\binom{n}{r}$, and $\binom{n}{r+1}$ for some integer $r$.

A crucial property of binomial coefficients is their ratio: $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k} \quad \text{and} \quad \frac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k}$

Step-by-step Derivation

We are given that three consecutive coefficients are in the ratio $2:5:12$. Let these be $C_{r-1}, C_r, C_{r+1}$ such that $C_{r-1} = \binom{n}{r-1}$, $C_r = \binom{n}{r}$, and $C_{r+1} = \binom{n}{r+1}$.

1. Set up the first ratio: The ratio of the first two consecutive coefficients is given as $2:5$. $\frac{C_{r-1}}{C_r} = \frac{2}{5}$ Substituting the binomial coefficient definitions: $\frac{\binom{n}{r-1}}{\binom{n}{r}} = \frac{2}{5}$ Using the ratio formula $\frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}$ (with $k=r$): $\frac{r}{n-r+1} = \frac{2}{5}$ To eliminate the denominators, we cross-multiply: $5r = 2(n-r+1)$ Distribute the 2 on the right side: $5r = 2n - 2r + 2$ Collect terms involving $r$ on one side: $5r + 2r = 2n + 2$ $7r = 2n + 2 \quad \text{(Equation 1)}$ This equation establishes a relationship between $n$ and $r$ based on the first given ratio.

2. Set up the second ratio: The ratio of the next two consecutive coefficients is given as $5:12$. $\frac{C_r}{C_{r+1}} = \frac{5}{12}$ Substituting the binomial coefficient definitions: $\frac{\binom{n}{r}}{\binom{n}{r+1}} = \frac{5}{12}$ Using the ratio formula $\frac{\binom{n}{k}}{\binom{n}{k+1}} = \frac{k+1}{n-k}$ (with $k=r$): $\frac{r+1}{n-r} = \frac{5}{12}$ To eliminate the denominators, we cross-multiply: $12(r+1) = 5(n-r)$ Distribute terms on both sides: $12r + 12 = 5n - 5r$ Collect terms involving $r$ on one side and $n$ on the other, or rearrange into standard form: $12r + 5r = 5n - 12$ $17r = 5n - 12 \quad \text{(Equation 2)}$ This equation provides a second relationship between $n$ and $r$.

3. Solve the system of linear equations: Now we have a system of two linear equations with two variables, $n$ and $r$:

1. $7r = 2n + 2$
2. $17r = 5n - 12$

From Equation 1, we can express $r$ in terms of $n$: $r = \frac{2n+2}{7}$ Substitute this expression for $r$ into Equation 2: $17\left(\frac{2n+2}{7}\right) = 5n - 12$ To eliminate the fraction, multiply the entire equation by 7: $17(2n+2) = 7(5n - 12)$ Distribute and simplify both sides: $34n + 34 = 35n - 84$ Now, collect terms involving $n$ on one side and constant terms on the other: $34 + 84 = 35n - 34n$ $118 = n$ Therefore, $n = 118$.

4. Verify the value of r: Since $n$ must be a positive integer, $n=118$ is a valid value. We can find $r$ using Equation 1: $7r = 2(118) + 2$ $7r = 236 + 2$ $7r = 238$ $r = \frac{238}{7}$ $r = 34$ Since $r=34$ is an integer, the coefficients $\binom{118}{33}$, $\binom{118}{34}$, $\binom{118}{35}$ are well-defined. The conditions for binomial coefficients require $0 \le r-1 < r < r+1 \le n$, which $33 < 34 < 35 \le 118$ satisfies.

Tips and Common Mistakes to Avoid

• Correct Ratio Formula: Always ensure you use the correct form of the ratio formula. A common mistake is to invert the ratio or incorrectly substitute $k$.
  • $\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k}$
  • $\frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}$
• Algebraic Precision: Be very careful with algebraic manipulations, especially when cross-multiplying and distributing terms. Small errors can lead to incorrect values for $n$ and $r$.
• Integer Check: Remember that $n$ must be a positive integer, and $r$ (the index of the term) must be a non-negative integer within the bounds of $n$ for the binomial coefficients to be valid.

Summary

By correctly applying the ratio property of consecutive binomial coefficients and carefully solving the resulting system of linear equations, we found the value of $n$. The problem highlights the importance of understanding the general term of a binomial expansion and the relationships between its coefficients. The value of $n$ that satisfies the given conditions is $118$.