Rewritten Solution

1. Key Concepts and Formulas

In the binomial expansion of $(1+x)^N$, the general term is given by $T_{r+1} = \binom{N}{r} x^r$, where $\binom{N}{r}$ represents the binomial coefficient.

Property of Binomial Coefficients in Arithmetic Progression (A.P.): If three consecutive binomial coefficients, $\binom{N}{r-1}$, $\binom{N}{r}$, and $\binom{N}{r+1}$, are in Arithmetic Progression, they satisfy the condition: $2 \binom{N}{r} = \binom{N}{r-1} + \binom{N}{r+1}$ This condition can be simplified using the ratio properties of binomial coefficients: $\frac{\binom{N}{k}}{\binom{N}{k-1}} = \frac{N-k+1}{k}$ Applying this to our A.P. condition by dividing by $\binom{N}{r}$, we get: $2 = \frac{\binom{N}{r-1}}{\binom{N}{r}} + \frac{\binom{N}{r+1}}{\binom{N}{r}}$ Using the ratio property, this transforms into a more direct formula for solving such problems: $2 = \frac{r}{N-r+1} + \frac{N-r}{r+1}$ This identity is particularly useful as it avoids extensive factorial manipulations.

Finding the Largest Binomial Coefficient: For an expansion of the form $(1+x)^M$:

• If $M$ is an even number, the largest binomial coefficient is the coefficient of the middle term, given by $\binom{M}{M/2}$.
• If $M$ is an odd number, there are two middle terms with equal largest coefficients, given by $\binom{M}{(M-1)/2}$ and $\binom{M}{(M+1)/2}$.

2. Step-by-Step Derivations

Given the binomial expansion $(1+x)^{n+4}$. Let $N$ be the exponent of the binomial, so $N = n+4$.

Step 1: Identify the coefficients. The 5th, 6th, and 7th terms in the expansion of $(1+x)^N$ correspond to $r=4$, $r=5$, and $r=6$ respectively in the general term $T_{r+1} = \binom{N}{r} x^r$. Therefore, their coefficients are:

• Coefficient of the 5th term ($T_5$): $C_5 = \binom{N}{4} = \binom{n+4}{4}$
• Coefficient of the 6th term ($T_6$): $C_6 = \binom{N}{5} = \binom{n+4}{5}$
• Coefficient of the 7th term ($T_7$): $C_7 = \binom{N}{6} = \binom{n+4}{6}$

Step 2: Apply the A.P. condition. Since these coefficients are in Arithmetic Progression, we use the fundamental A.P. property: $2 \times (\text{middle term}) = (\text{first term}) + (\text{last term})$ Substituting our coefficients: $2 \binom{n+4}{5} = \binom{n+4}{4} + \binom{n+4}{6}$

Step 3: Simplify the equation using the ratio property of binomial coefficients. Instead of expanding factorials, we employ the simplified identity for coefficients in A.P.: $2 = \frac{r}{N-r+1} + \frac{N-r}{r+1}$ In our context, $N = n+4$ and the middle term's $r$ value is $5$. Substituting these values into the identity: $2 = \frac{5}{(n+4)-5+1} + \frac{(n+4)-5}{5+1}$ Simplifying the denominators: $2 = \frac{5}{n+0} + \frac{n-1}{6}$ $2 = \frac{5}{n} + \frac{n-1}{6}$ To clear the denominators, multiply the entire equation by $6n$ (the least common multiple of $n$ and $6$): $2 \times 6n = \left(\frac{5}{n}\right) \times 6n + \left(\frac{n-1}{6}\right) \times 6n$ $12n = 30 + n(n-1)$ $12n = 30 + n^2 - n$ Rearrange the terms to form a standard quadratic equation: $n^2 - n - 12n + 30 = 0$ $n^2 - 13n + 30 = 0$

Step 4: Solve the quadratic equation for $n$. We can solve this quadratic equation by factorization. We look for two numbers that multiply to $30$ and add up to $-13$. These numbers are $-3$ and $-10$. $(n-3)(n-10) = 0$ This yields two possible values for $n$: $n=3 \quad \text{or} \quad n=10$

Step 5: Validate the value of $n$. The problem statement explicitly mentions the condition $n \neq 10$. Therefore, we must reject $n=10$. The only valid value for $n$ is $n=3$.

Step 6: Determine the largest coefficient in the expansion. Now that we have $n=3$, we can find the actual exponent of the binomial expansion $(1+x)^{n+4}$. The exponent is $N = n+4 = 3+4 = 7$. So, we need to find the largest coefficient in the expansion of $(1+x)^7$. Here, the exponent $M=7$, which is an odd number. As per the rule, for an odd exponent, the largest coefficients are the coefficients of the middle terms. The middle terms correspond to $r = (M-1)/2$ and $r = (M+1)/2$. For $M=7$:

• $r_1 = (7-1)/2 = 3$
• $r_2 = (7+1)/2 = 4$ The largest coefficients are $\binom{7}{3}$ and $\binom{7}{4}$. These values are equal due to the symmetry property of binomial coefficients, $\binom{N}{k} = \binom{N}{N-k}$. Let's calculate the value of $\binom{7}{3}$: $\binom{7}{3} = \frac{7!}{3!(7-3)!} = \frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35$ Thus, the largest coefficient in the expansion of $(1+x)^{7}$ is $35$.

3. Tips and Common Mistakes to Avoid

• Misidentifying Coefficients: Ensure you correctly identify the 'r' value for the $k^{th}$ term. Remember that the $k^{th}$ term is $T_k$, which corresponds to $\binom{N}{k-1}$. For the 5th term, $r=4$.
• Ignoring Conditions: Always check for explicit conditions in the problem, like $n \neq 10$, as they are critical for choosing the correct solution from multiple possibilities.
• Calculation Errors: Be careful with algebraic manipulation when solving the quadratic equation.
• Largest Coefficient for Even vs. Odd Exponents: Differentiate between how to find the largest coefficient when the exponent $M$ is even versus when it is odd.

4. Summary and Key Takeaway

This problem effectively tests the understanding of binomial coefficients and their properties, especially when they form an Arithmetic Progression. The most efficient approach involves using the derived ratio identity $2 = \frac{r}{N-r+1} + \frac{N-r}{r+1}$ to form a quadratic equation for $n$. After solving for $n$ and validating it against given conditions, the final step requires knowing how to find the largest binomial coefficient for a given exponent. The key takeaway is the strategic application of identities to simplify calculations and careful attention to problem constraints.