For some n 10, let the coefficients of the 5 th, 6 th and 7 th terms in the binomial expan…

Question

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

Accepted Answer

$$\begin{aligned}
& (1+\mathrm{x})^{\mathrm{n+4}} \
& { }^{\mathrm{n}+4} \mathrm{C}_4{ }^{\mathrm{n+4}} \mathrm{C}_5{ }^{\mathrm{n}+4} \mathrm{C}_6, \rightarrow \text { A.P. } \
& \Rightarrow 2 \times{ }^{\mathrm{n}+4} \mathrm{C}_5={ }^{\mathrm{n+4}} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_6 \
& \Rightarrow 4 \times{ }^{\mathrm{n+4}} \mathrm{C}_5=\left({ }^{n+4} \mathrm{C}_4+{ }^{\mathrm{n}+4} \mathrm{C}_5\right)+\left({ }^{\mathrm{n+4}} \mathrm{C}_5+{ }^{\mathrm{n}+4} \mathrm{C}_6\right) \
& \Rightarrow 4 \times{ }^{n+4} \mathrm{C}_5={ }^{\mathrm{n}+5} \mathrm{C}_5+{ }^{\mathrm{n}+5} \mathrm{C}_6 \
& \Rightarrow 4 \times \frac{(\mathrm{n}+4)!}{5!.(\mathrm{n}-1)!}=\frac{(\mathrm{n}+6)!}{6!\mathrm{n}!} \
& \Rightarrow 4=\frac{(\mathrm{n}+6)(\mathrm{n}+5)}{6 \mathrm{n}} \
& \Rightarrow \mathrm{n}^2+11 \mathrm{n}+30=24 \mathrm{n} \
& \Rightarrow \mathrm{n}^2-13 \mathrm{n}+30=0 \
& \Rightarrow \mathrm{n}=3,10(\text { rejected }) \
& \because \mathrm{n} \neq 10
\end{aligned}$$

$\therefore$ Largest binomial coefficient in expansion of

$$\begin{aligned}
& (1+x)^7 \
& (\because \mathrm{n}+4=7)
\end{aligned}$$

is coeff. of middle term

$$\Rightarrow{ }^7 \mathrm{C}_4={ }^7 \mathrm{C}_3=35$$

N.T.A. Ans Option (2)

Solutions · 1