Find the only positive integer $n$ such that row $n$ of Pascal's Triangle contains three successive entries with the ratio $5:6:7.$

We are tasked with finding the only positive integer $ n $ such that row $ n $ of Pascal's Triangle contains three successive entries with the ratio $ 5:6:7 $.

### Step 1: Review of Pascal's Triangle
The entries in row $ n $ of Pascal's Triangle are the binomial coefficients $ \binom{n}{0}, \binom{n}{1}, \binom{n}{2}, \dots, \binom{n}{n} $. Our goal is to find three successive entries from this row that are in the ratio $ 5:6:7 $. That is, we are looking for three consecutive binomial coefficients $ \binom{n}{k-1}, \binom{n}{k}, \binom{n}{k+1} $ such that:

\[
\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{6}{5} \quad \text{and} \quad \frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{7}{6}.
\]

### Step 2: Express the Ratios Using Binomial Coefficients
Using the properties of binomial coefficients, we know that:

\[
\frac{\binom{n}{k}}{\binom{n}{k-1}} = \frac{n-k+1}{k} \quad \text{and} \quad \frac{\binom{n}{k+1}}{\binom{n}{k}} = \frac{n-k}{k+1}.
\]

We are given that:

\[
\frac{n-k+1}{k} = \frac{6}{5} \quad \text{and} \quad \frac{n-k}{k+1} = \frac{7}{6}.
\]

### Step 3: Solve the System of Equations
We now solve the system of equations:

1. $\frac{n-k+1}{k} = \frac{6}{5}$
2. $\frac{n-k}{k+1} = \frac{7}{6}$

#### Equation 1:
From $\frac{n-k+1}{k} = \frac{6}{5}$, we can rewrite this as:

\[
n-k+1 = \frac{6}{5}k \implies 5(n-k+1) = 6k \implies 5n - 5k + 5 = 6k \implies 5n + 5 = 11k \implies k = \frac{5n + 5}{11}.
\]

#### Equation 2:
From $\frac{n-k}{k+1} = \frac{7}{6}$, we can rewrite this as:

\[
n-k = \frac{7}{6}(k+1) \implies 6(n-k) = 7(k+1) \implies 6n - 6k = 7k + 7 \implies 6n = 13k + 7 \implies k = \frac{6n - 7}{13}.
\]

### Step 4: Equate the Two Expressions for $ k $
We now have two expressions for $ k $:

\[
\frac{5n + 5}{11} = \frac{6n - 7}{13}.
\]

Cross-multiply to solve for $ n $:

\[
13(5n + 5) = 11(6n - 7) \implies 65n + 65 = 66n - 77 \implies 65 + 77 = 66n - 65n \implies n = 142.
\]

### Step 5: Find $ k $
Substitute $ n = 142 $ into either expression for $ k $. Using $ k = \frac{5n + 5}{11} $:

\[
k = \frac{5(142) + 5}{11} = \frac{710 + 5}{11} = \frac{715}{11} = 65.
\]

### Step 6: Verify the Solution
We now verify that the binomial coefficients $ \binom{142}{64}, \binom{142}{65}, \binom{142}{66} $ are in the ratio $ 5:6:7 $.

1. The ratio $\frac{\binom{142}{65}}{\binom{142}{64}} = \frac{142-64+1}{65} = \frac{79}{65} = \frac{6}{5}$.
2. The ratio $\frac{\binom{142}{66}}{\binom{142}{65}} = \frac{142-65}{66} = \frac{77}{66} = \frac{7}{6}$.

Both ratios are correct, confirming that $ n = 142 $ is the solution.

### Final Answer:
The only positive integer $ n $ such that row $ n $ of Pascal's Triangle contains three successive entries with the ratio $ 5:6:7 $ is $ \boxed{142} $.

---

Would you like further details or clarifications on any part? Here are five related questions to deepen your understanding:

1. What are the properties of binomial coefficients used to derive the ratios?
2. How do the combinatorial properties of Pascal's Triangle help in finding patterns like these?
3. What happens if the ratio was different, say $ 3:4:5 $?
4. How can the solution method change for non-consecutive entries in Pascal’s Triangle?
5. How do these ratios relate to approximations of Pascal's Triangle for large $ n $?

**Tip:** Understanding the recursive nature of binomial coefficients helps in solving complex combinatorial problems effectively.

Explore how to find the only row in Pascal's Triangle where three consecutive entries are in the ratio 5:6:7. This combinatorics problem involves solving a system of equations using binomial coefficients. Ideal for undergraduate students studying advanced mathematics.

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