Elaborate Solution to Binomial Coefficient Problem

Key Concept: Ratio of Consecutive Binomial Coefficients

For a binomial expansion of the form $(a+b)^N$, the ratio of two consecutive binomial coefficients, $\binom{N}{k}$ and $\binom{N}{k+1}$, is given by the formula: $\frac{\binom{N}{k}}{\binom{N}{k+1}} = \frac{k+1}{N-k}$ This fundamental property is crucial for problems involving ratios of consecutive terms in a binomial expansion, as it allows us to establish relationships between the exponent $N$ and the term index $k$.

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Step-by-Step Solution

1. Identifying the Coefficients and Their Ratios The given binomial expansion is $(1 + x)^{n + 5}$. Let $N = n+5$ be the total exponent. Let the three consecutive terms in the expansion have coefficients $^{N}C_r$, $^{N}C_{r+1}$, and $^{N}C_{r+2}$. According to the problem, their ratio is $5 : 10 : 14$. This gives us two pairs of ratios:

• The ratio of the first two coefficients: $\frac{^{N}C_r}{^{N}C_{r+1}} = \frac{5}{10} = \frac{1}{2}$
• The ratio of the second and third coefficients: $\frac{^{N}C_{r+1}}{^{N}C_{r+2}} = \frac{10}{14} = \frac{5}{7}$

2. Formulating Equations using the Ratio Formula

• For the first ratio: We apply the key concept with $N = n+5$ and $k=r$. $\frac{^{n+5}C_r}{^{n+5}C_{r+1}} = \frac{r+1}{(n+5)-r}$ Since this ratio is equal to $1/2$, we set up the equation: $\frac{r+1}{n+5-r} = \frac{1}{2}$ Explanation: We equate the derived ratio using the formula to the given ratio ($1/2$) to form an algebraic equation involving $n$ and $r$. Cross-multiplying gives: $2(r+1) = 1(n+5-r)$ $2r+2 = n+5-r$ Rearranging the terms to isolate $n$: $3r = n+3 \quad \ldots(i)$

• For the second ratio: Similarly, we apply the key concept with $N = n+5$ and $k=r+1$. $\frac{^{n+5}C_{r+1}}{^{n+5}C_{r+2}} = \frac{(r+1)+1}{(n+5)-(r+1)}$ $= \frac{r+2}{n+5-r-1}$ $= \frac{r+2}{n+4-r}$ Since this ratio is equal to $5/7$, we set up the equation: $\frac{r+2}{n+4-r} = \frac{5}{7}$ Explanation: Just like the first ratio, we use the formula and equate it to the given ratio ($5/7$) to obtain another equation linking $n$ and $r$. Cross-multiplying gives: $7(r+2) = 5(n+4-r)$ $7r+14 = 5n+20-5r$ Rearranging the terms: $12r = 5n+6 \quad \ldots(ii)$

3. Solving the System of Equations

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

1. $n = 3r - 3$ (from equation (i))
2. $12r = 5n + 6$ (from equation (ii))

Explanation: We will use the substitution method to solve for $n$ and $r$. This involves expressing one variable in terms of the other from one equation and substituting it into the second equation. Substitute the expression for $n$ from equation (i) into equation (ii): $12r = 5(3r-3) + 6$ $12r = 15r - 15 + 6$ $12r = 15r - 9$ Now, we solve for $r$: $9 = 15r - 12r$ $9 = 3r$ $r = 3$

Now that we have the value of $r$, substitute it back into equation (i) to find $n$: $n = 3(3) - 3$ $n = 9 - 3$ $n = 6$

4. Determining the Binomial Expansion and Total Power

We found $n=6$. The original binomial expansion was $(1+x)^{n+5}$. Substituting $n=6$, the total exponent of the expansion is $N = 6+5 = 11$. So, the expansion is $(1+x)^{11}$.

5. Finding the Largest Coefficient

For a binomial expansion $(a+b)^N$, the largest coefficient occurs at the middle term(s).

• If $N$ is even, the largest coefficient is $\binom{N}{N/2}$.
• If $N$ is odd, there are two equal largest coefficients: $\binom{N}{(N-1)/2}$ and $\binom{N}{(N+1)/2}$.

In our case, the exponent $N=11$, which is an odd number. Therefore, the largest coefficients are $^{11}C_{(11-1)/2}$ and $^{11}C_{(11+1)/2}$. This means the largest coefficients are $^{11}C_5$ and $^{11}C_6$. Since $^{N}C_k = ^{N}C_{N-k}$, we know that $^{11}C_5 = ^{11}C_{11-5} = ^{11}C_6$. Both will have the same value.

Let's calculate $^{11}C_6$: $^{11}C_6 = \frac{11!}{6!(11-6)!} = \frac{11!}{6!5!}$ $= \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6!}{6! \times 5 \times 4 \times 3 \times 2 \times 1}$ $= \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1}$ $= 11 \times (10/5/2) \times (9/3) \times (8/4) \times 7$ $= 11 \times 1 \times 3 \times 2 \times 7$ $= 462$

Thus, the largest coefficient in the expansion is 462.

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Tips for Success / Common Mistakes

• Correct Indexing: Pay close attention to the index $k$ when applying the ratio formula. If the coefficients are $^{N}C_k$ and $^{N}C_{k+1}$, the formula is $\frac{k+1}{N-k}$. Ensure you correctly identify which index is $k$.
• Algebraic Precision: Double-check your algebraic manipulations when solving the system of equations. A small error can lead to incorrect values for $n$ and $r$.
• Largest Coefficient Rule: Remember the distinct rules for finding the largest coefficient based on whether the binomial exponent $N$ is even or odd. If $N$ is odd, there are two numerically equal largest coefficients.
• Context of $N$: In this problem, the exponent is $n+5$, not just $n$. Ensure you correctly use $n+5$ as the $N$ in your calculations.

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Summary / Key Takeaway

This problem demonstrates the powerful application of the ratio of consecutive binomial coefficients to determine unknown parameters ($n$ and $r$) within a binomial expansion. Once these parameters are found, identifying the largest coefficient involves applying the rule for the middle term(s) of the expansion. Precision in applying the formulas and careful algebraic calculation are key to arriving at the correct solution.