Question

If a, b, c and d in any binomial expansion be the 6th, 7th, 8th and 9th terms respectively, then prove that $\frac{b^2-ac}{c^2-bd}=\frac{4a}{3c}$.

Answer 1

$$\begin{gathered} Supposethebinomialexpressionis(1+x{)}^n. \\ Then,the6th,7th,8thand9thtermsare{}^{n}C_5{x}^5,{}^{n}C_6{x}^6,{}^{n}C_7{x}^{7}and{}^{n}C_8{x}^8,respectively. \\ Now,wehave: \\ \frac{{}^{n}C_{6x^6}}{{}^{n}C_5{x}^5}=\frac{b}{a},\frac{{}^{n}C_{8x^8}}{{}^{n}C_7{x}^7}=\frac{d}{c}and\frac{{}^{n}C_{7x^7}}{{}^{n}C_6{x}^6}=\frac{c}{b} \\ \Rightarrow \frac{n-5}{6}=\frac{b}{a}\mathrm{and}\frac{n-6}{7}=\frac{c}{b} \\ \Rightarrow \frac{{\displaystyle \raisebox{1ex}{$n-6$}\!\left/ \!\raisebox{-1ex}{$7$}\right.}}{{\displaystyle \raisebox{1ex}{$n-5$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}}=\frac{{\displaystyle \raisebox{1ex}{$c$}\!\left/ \!\raisebox{-1ex}{$b$}\right.}}{{\displaystyle \raisebox{1ex}{$b$}\!\left/ \!\raisebox{-1ex}{$a$}\right.}} \\ \Rightarrow \frac{6n-36}{7n-35}=\frac{c}{a} \end{gathered}$$