For some n neq 10$n \neq 10$, let the coefficients of the 5^textth$5^{\text{th}}$, 6^textth$6^{\text{th}}$ and 7^textth$7^{\text{th}}$ terms in the binomial expansion of (1 + x)^n+4$(1 + x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of (1 + x)^n+4$(1 + x)^{n+4}$ is :

### Related Formula The coefficient of the r^textth$r^{\text{th}}$ term in the expansion (1+x)^m$(1+x)^m$ is written as binommr-1$\binom{m}{r-1}$. For three terms in A.P., their values satisfy: 2 cdot T_2 = T_1 + T_3$$2 \cdot T_2 = T_1 + T_3$$ ### Core Logic Let the total power exponent be m = n + 4$m = n + 4$. The coefficients of the 5^textth$5^{\text{th}}$, 6^textth$6^{\text{th}}$, and 7^textth$7^{\text{th}}$ terms are binomm4$\binom{m}{4}$, binomm5$\binom{m}{5}$, and binomm6$\binom{m}{6}$ respectively. Since they form an arithmetic progression: 2 cdot binomm5 = binomm4 + binomm6$$2 \cdot \binom{m}{5} = \binom{m}{4} + \binom{m}{6}$$ Rearrange using the recurrence addition identity properties: 4 cdot binomm5 = left[ binomm4 + binomm5 right] + left[ binomm5 + binomm6 right]$$4 \cdot \binom{m}{5} = \left[ \binom{m}{4} + \binom{m}{5} \right] + \left[ \binom{m}{5} + \binom{m}{6} \right]$$ 4 cdot binomm5 = binomm+15 + binomm+16$$4 \cdot \binom{m}{5} = \binom{m+1}{5} + \binom{m+1}{6}$$ 4 cdot binomm5 = binomm+26$$4 \cdot \binom{m}{5} = \binom{m+2}{6}$$ ### Step 1: Solve the Combinatorial Fraction for m$m$ Expand the combinations using factorials: 4 cdot fracm!5!(m-4)! = frac(m+2)!6!(m-4)!$$4 \cdot \frac{m!}{5!(m-4)!} = \frac{(m+2)!}{6!(m-4)!}$$ Cancel out (m-4)!$(m-4)!$ from both denominators: 4 cdot fracm!120 = frac(m+2)(m+1)m!720$$4 \cdot \frac{m!}{120} = \frac{(m+2)(m+1)m!}{720}$$ 4 = frac(m+2)(m+1)6$$4 = \frac{(m+2)(m+1)}{6}$$ 24 = m^2 + 3m + 2 implies m^2 + 3m - 22 = 0$$24 = m^2 + 3m + 2 \implies m^2 + 3m - 22 = 0$$ Wait, let's re-verify the step using the direct ratio computation: 2 = fracbinomm4binomm5 + fracbinomm6binomm5 = frac5m-4 + fracm-56$$2 = \frac{\binom{m}{4}}{\binom{m}{5}} + \frac{\binom{m}{6}}{\binom{m}{5}} = \frac{5}{m-4} + \frac{m-5}{6}$$ 2 = frac30 + (m-4)(m-5)6(m-4)$$2 = \frac{30 + (m-4)(m-5)}{6(m-4)}$$ 12(m-4) = 30 + m^2 - 9m + 20$$12(m-4) = 30 + m^2 - 9m + 20$$ 12m - 48 = m^2 - 9m + 50 implies m^2 - 21m + 98 = 0$$12m - 48 = m^2 - 9m + 50 \implies m^2 - 21m + 98 = 0$$ Factor the quadratic equation: (m-7)(m-14) = 0 implies m = 7 text or m = 14$$(m-7)(m-14) = 0 \implies m = 7 \text{ or } m = 14$$ ### Step 2: Connect back to the problem constraints Since m = n + 4$m = n + 4$: - If m = 14 implies n + 4 = 14 implies n = 10$m = 14 \implies n + 4 = 14 \implies n = 10$ (this is rejected because the problem states n neq 10$n \neq 10$). - If m = 7 implies n + 4 = 7 implies n = 3$m = 7 \implies n + 4 = 7 \implies n = 3$ (this value is accepted). Thus, the binomial expansion exponent is exactly m = 7$m = 7$. ### Step 3: Extract Maximum Binomial Coefficient For an odd exponent power m = 7$m = 7$, the maximum binomial coefficient corresponds to the middle terms: textMax Coefficient = binom73 = binom74 = frac7 cdot 6 cdot 53 cdot 2 cdot 1 = 35$$\text{Max Coefficient} = \binom{7}{3} = \binom{7}{4} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$$ ### Pattern Recognition For consecutive binomial coefficients binomnr-1, binomnr, binomnr+1$\binom{n}{r-1}, \binom{n}{r}, \binom{n}{r+1}$ in A.P., the power parameters satisfy the standard identity (n-2r)^2 = n+2$(n-2r)^2 = n+2$, which allows quick calculation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Binomial Theorem