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Mathematics Question on Binomial theorem

Let the coefficients of the middle terms in the expansion of(16+βx)4,(1–3βx)2and(1–β2x)6,β>0,\begin{array}{l} \left(\dfrac{1}{\sqrt 6}+ \beta x\right)^4, \left(1 – 3\beta x\right)^2 \text{and}\left(1 – \dfrac{\beta}{2}x\right )^6, \beta > 0,\end{array}respectively form the first three terms of an A.P. If d is the common difference of this A.P., then 50−2dβ2\begin{array}{l} 50-\frac{2d}{\beta^2} \end{array}is equal to _______.

Answer

Coefficients of middle terms

4C216β2, 2C1(−3β), 6C3(−β2)3form an A.P.∴ 2.2(−3β)=β2−5β32\begin{array}{l} ^4C_2\frac{1}{6}\beta^2,\ ^2C_1\left(-3\beta\right),\ ^6C_3\left(\frac{-\beta}{2}\right)^3 \text{form an A.P.} \\\\\therefore\ 2.2\left(-3\beta\right)=\beta^2-\frac{5\beta^3}{2}\end{array}

⇒−24=2β–5β2 ⇒5β2−2β−24=0 ⇒5β2−12β+10β–24=0 ⇒β(5β−12)+2(5β–12)=0\begin{array}{l}\Rightarrow -24 = 2\beta – 5\beta^2 \\\ \Rightarrow 5\beta^2- 2\beta- 24 = 0\\\ \Rightarrow 5\beta^2- 12\beta + 10\beta – 24 = 0\\\ \Rightarrow \beta \left(5\beta- 12\right) + 2 \left(5\beta – 12\right) = 0\end{array}

β=125 d=−6β–β2\begin{array}{l}\beta = \frac{12}{5} \\\ d = -6\beta – \beta^2\end{array}

∴ 50−2dβ2=50−2(−6β−β2)β2=50+12β+2=57\begin{array}{l} \therefore\ 50-\frac{2d}{\beta^2}=50-2\frac{\left(-6\beta-\beta^2\right)}{\beta^2}=50+\frac{12}{\beta}+2=57\end{array}