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

Let

$T_6=a+{}^n{C}_5{x}^{n-5}{\alpha }^5$

$=\frac{n!}{5!(n-5)!}\times x^{n-5}{\alpha }^5=\frac{n(n-1)(n-2)(n-3)(n-4)}{120}\times x^{n-5}{\alpha }^5$

$T_7=b={}^n{C}_6{x}^{n-6}{\alpha }^6$

$=\frac{n!}{6!(n-6)!}\times x^{n-r}{\alpha }^6=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{720}\times x^{n-6}{\alpha }^6$

$T_8=c={}^n{C}_7{x}^{n-7}{\alpha }^7$

$=\frac{n!}{7!(n-7)!}\times x^{n-7}{\alpha }^7=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{5040}\times x^{n-7}{\alpha }^7$

$T_9=d={}^n{C}_8{x}^{n-8}{\alpha }^8$

$=\frac{n!}{8!(n-8)!}\times x^{n-8}{\alpha }^8=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{40320}\times x^{n-8}{\alpha }^8$

To prove: $\frac{b^2-ac}{c^2-bd}=\frac{4a}{3c}$

Now dividing b by a i.e. $T_7$ by $T_6$

$\frac{T_7}{T_6}=\frac{b}{a}=\frac{\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{720}\times x^{n-6}{\alpha }^6}{\frac{n(n-1)(n-2)(n-3)(n-4)}{120}\times x^{n-5}{\alpha }^5}$

$=\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)\times 120\times x^{n-6}{\alpha }^6}{720n(n-1)(n-2)(n-3)(n-4)\times x^{n-5}{\alpha }^5}=\frac{n-5}{6}\times x^{-1}\alpha$

$=\frac{(n-5)\alpha }{6x}....\left(1\right)$

Now,

$\frac{T_8}{T_7}=\frac{c}{d}=\frac{\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{5040}}{\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)}{720}}\times x^{n-6}{\alpha }^6$

$=\frac{(n-6)}{5040}\times 720\times x^{-1}\alpha =\frac{(n-6)}{7x}\times \alpha ....\left(2\right)$

and

$\frac{T_9}{T_8}=\frac{d}{c}=\frac{\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)(n-7)}{40320}\times x^{n-8}{\alpha }^8}{\frac{n(n-1)(n-2)(n-3)(n-4)(n-5)(n-6)}{5040}\times x^{n-7}{\alpha }^7}=\frac{(n-6)}{40320}\times 5040\times x^{-1}\times \alpha =\frac{(n-7)}{8x}\alpha .....\left(3\right)$

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

$=\frac{\frac{b}{a}}{\frac{c}{a}}=\frac{\frac{(n-5)\alpha }{6x}}{\frac{(n-6)}{7x}\times \alpha }=\frac{b^2}{ac}=\frac{(n-5)}{6}\times \frac{7}{(n-6)}$

$=\frac{b^2}{ac}=\frac{7(n-5)}{6(n-6)}.....\left(4\right)$

and

$=\frac{\frac{c}{b}}{\frac{d}{c}}=\frac{\frac{(n-6)}{7x}\times \alpha }{\frac{(n-7)}{8x}\times \alpha }$

$=\frac{c^2}{bd}=\frac{8(n-6)}{7(n-7)}$

Now subtract 1 from both sides of equation

$=\frac{b^2}{ac}-1=\frac{7(n-5)}{6(n-6)}-1$

$\Rightarrow \frac{b^2-ac}{ac}=\frac{7n-35-6n+36}{6(n-6)}=\frac{n+1}{6(n-6)}......\left(6\right)$

and

$\frac{c^2}{bd}-1=\frac{8(n-6)}{7(n-7)}-1=\frac{8n-48-7n+49}{7(n-7)}=\frac{c-bd}{bd}=\frac{(n+1)}{7(n-7)}....\left(7\right)$

Again, on dividing 6 by 7

$\frac{\frac{b^2-ac}{ac}}{\frac{c^2-bd}{bd}}=\frac{\frac{(n+1)}{6(n-6)}}{\frac{(n+1)}{7(n-7)}}=\frac{b^2-ac}{c^2-bd}\times \frac{bd}{ac}=\frac{7(n-7)}{6(n-6)}.......\left(8\right)$

On multiplying (5) by (8)

$=\frac{c^2-bd}{b^2-ac}\times \frac{ac}{bd}\times \frac{c^2}{bd}=\frac{48(n-6{)}^2}{49(n-7{)}^2}=\frac{c^2}{bd}\times \frac{b^2-ac}{c^2-bd}\times \frac{bd}{ac}=\frac{8(n-6)}{7(n-7)}\times \frac{7(n-7)}{6(n-6)}$

$=\frac{b^2-ac}{c^2-bd}\times \frac{c}{a}=\frac{4}{3}=\frac{b^2-ac}{c^2-bd}=\frac{4a}{3c}$

Hence proved.