Binomial Expansion

Question

In the expansion of (1 + x)ⁿ with n > 10, if the coefficients of the 5th, 6th, and 7th terms are in AP, then n equals

RankGuru · Accepted Answer

1. **Identify Coefficients**: The coefficients of $5^{th}, 6^{th}, 7^{th}$ terms are $\binom{n}{4}, \binom{n}{5}, \binom{n}{6}$.
2. **Arithmetic Progression Condition**: $2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6}$.
3. **Divide by $\binom{n}{5}$**: $2 = \frac{\binom{n}{4}}{\binom{n}{5}} + \frac{\binom{n}{6}}{\binom{n}{5}}$.
   - Use property $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} \implies \frac{\binom{n}{5}}{\binom{n}{4}} = \frac{n-4}{5}$ and $\frac{\binom{n}{6}}{\binom{n}{5}} = \frac{n-5}{6}$.
4. **Equation**: $2 = \frac{5}{n-4} + \frac{n-5}{6} \implies 12 = \frac{30}{n-4} + (n-5)$
   - Solving $n^2 - 21n + 98 = 0 \implies (n-7)(n-14) = 0$.
5. **Condition Check**: Given $n > 10$, we take **$n = 14$**.

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