If the coefficients of x^4, x^5 and x^6 in the expansion of (1+x)^n are in the arithmetic …

Question

If the coefficients of $x^4, x^5$ and $x^6$ in the expansion of $(1+x)^n$ are in the arithmetic progression, then the maximum value of $n$ is:

Accepted Answer

$\begin{gathered} (1+x{)}^{n}={}^{n}{C}_{0}+{}^{n}{C}_{1}{x}^{1}+{}^{n}{C}_{2}{x}^{2}+\ldots {}^{n}{C}_{n}{x}^{n} \ {}^{n}{C}_{4},{}^{n}{C}_{5}{}^{n}{C}_{6} \ {}^{n}{C}_{5}-{}^{n}{C}_{4}={}^{n}{C}_{6}-{}^{n}{C}_{5} \ \Rightarrow \frac{n!}{5!(n-5)!}-\frac{n!}{4!(n-4)!}=\frac{n!}{6!(n-6)!}-\frac{n!}{5!(n-5)!} \ \Rightarrow 30(n-9)(n-6)=5(n-4)(n-11) \ \Rightarrow 30{n}^{2}-450n+1620=5{n}^{2} \ \Rightarrow \frac{1}{n-5}[\frac{n-4-5}{5(n-4)}]=\frac{1}{5}[\frac{n-5-6}{6(n-5)}] \ \Rightarrow \frac{n-9}{5(n-4)}=\frac{1}{5}[\frac{n-11}{6}] \ \Rightarrow {n}^{2}-21n+98=0 \ {n}_{max}=14 \end{gathered}$

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