Question

If 3rd, 4th, 5th and 6th terms in the expansion of $(x+\alpha {)}^n$ be respectively a,b,c and d. prove that $\frac{b^2-ac}{c^2-bd}=\frac{5a}{3c}.$

Answer 1

Given expansion is $(x+\alpha {)}^n$

Then,

$T_3=a={}^n{C}_2{x}^{n-2}{\alpha }^2=\frac{n(n-1)}{2}{x}^{n-2}{\alpha }^2$

$T_4=b={}^n{C}_3{x}^{n-3}{\alpha }^3=\frac{n(n-1)(n-2)}{6}{x}^{n-3}{\alpha }^3$

$T_5=c={}^n{C}_4{x}^{n-4}{\alpha }^4=\frac{n(n-1)(n-2)(n-3)}{24}{x}^{n-4}{\alpha }^4$

$T_6=d={}^n{C}_5{x}^{n-5}{\alpha }^5=\frac{n(n-1)(n-2)(n-3)(n-4)}{120}{x}^{n-5}{\alpha }^5$

Now

$\frac{T_4}{T_3}=\frac{b}{a}=\frac{\frac{n(n-1)(n-2)}{6}{x}^{n-3}{\alpha }^3}{\frac{n(n-1)}{2}{x}^{n-2}{\alpha }^2}=\left(\frac{n-2}{3}\right)\frac{\alpha }{x}....\left(1\right)$

Similarly,

$\frac{T_5}{T_4}=\frac{c}{b}=\frac{\frac{n(n-1)(n-2)(n-3)}{24}{x}^{n-4}{\alpha }^4}{\frac{n(n-1)(n-2)}{6}{x}^{n-3}{\alpha }^3}=\left(\frac{n-3}{4}\right)\frac{\alpha }{x}.......\left(2\right)$

and

$\frac{T_6}{T_5}=\frac{d}{c}=\frac{\frac{n(n-1)(n-2)(n-3)(n-4)}{120}}{\frac{n(n-1)(n-2)(n-3)}{24}{x}^{n-4}{\alpha }^4}=\left(\frac{n-4}{5}\right).\frac{\alpha }{x}........\left(3\right)$

Again, dividing (1)by (2) and (2)by (3), we get

$=\frac{\frac{b}{a}}{\frac{c}{a}}=\frac{\left(\frac{n-2}{3}\right).\frac{\alpha }{x}}{\left(\frac{n-3}{4}\right).\frac{\alpha }{x}}=\frac{4(n-2)}{3(n-3)}$

$=\frac{b^2}{ac}=\frac{4(n-2)}{3(n-3)}.....\left(4\right)$

and

$=\frac{\frac{c}{b}}{\frac{d}{c}}=\frac{\left(\frac{n-3}{4}\right).\frac{\alpha }{x}}{\left(\frac{n-4}{5}\right).\frac{\alpha }{x}}=\frac{5(n-3)}{4(n-4)}=\frac{c^2}{bc}=\frac{5(n-3)}{4(n-4)}........\left(5\right)$

Now subtact 1 from both sides of equation (4) and (5)as:

$=\frac{b^2}{ac}-1=\frac{4(n-2)}{3(n-3)}-1$

$\Rightarrow \frac{b^2-ac}{ac}=\frac{n+1}{3(n-3)}.....\left(6\right)$

and

$\frac{c^2}{bd}-1=\frac{5(n-3)}{4(n-4)}-1$

$\frac{c^2-bd}{bd}=\frac{n+1}{4(n-4)}......\left(7\right)$

Again, on dividing (6) by (7), we get

$\frac{\frac{b^2-ac}{ac}}{\frac{c^2-bd}{bd}}=\frac{\frac{(n+1)}{3(n-3)}}{\frac{(n+1)}{4(n-4)}}$

$\frac{b^2-ac}{c^2-bd}\times \frac{bd}{ac}=\frac{4(n-4)}{3(n-3)}.....\left(8\right)$

On multiplying (5)by (8)

$\Rightarrow \frac{b^2-ac}{c^2-bd}\times \frac{bd}{ac}\times \frac{c^2}{bd}=\frac{4(n-4)}{3(n-3)}\times \frac{5(n-3)}{4(n-4)}\Rightarrow \frac{b^2-ac}{c^2-bd}.\frac{c}{a}=\frac{5}{3}\Rightarrow \frac{b^2-ac}{c^2-bd}=\frac{5a}{3c}$

Hence, proved.