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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & (1+x)^{n+4} \ & { }^{n+4} C_4,{ }^{n+4} C_5,{ }^{n+4} C_6, ightarrow 	ext { A.P. } \ & \Rightarrow 2 	imes 	imes{ }^{n+4} C_5={ }^{n+4} C_4+{ }^{n+4} C_6 \ & \Rightarrow 4 	imes{ }^{n+4} C_5=\left({ }^{n+4} C_4+{ }^{n+4} C_5ight)+\left({ }^{n+4} C_5+{ }^{n+4} C_6ight) \ & \Rightarrow 4 	imes{ }^{n+4} C_5={ }^{n+5} C_5+{ }^{n+5} C_6\end{aligned}$ $\begin{aligned} & \Rightarrow 4 	imes \frac{(\mathrm{n}+4)!}{5!\cdot(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\cdot \mathrm{n}!} \ & \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \ & \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \ & \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \ & \Rightarrow \mathrm{n}=3,10(	ext { rejected }) \ & \because \mathrm{n} 
eq 10\end{aligned}$ $	herefore$ Largest binomial coefficient in expansion of $\begin{aligned} & (1+\mathrm{x})^7 \ & (\because \mathrm{n}+4=7) \end{aligned}$ is coeff. of middle term $\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35$

For some $n \neq 10$, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expansion of…

Solution