Question

The ratio of $(r+1{)}^{th}$ and $(r-1{)}^{th}$ terms in the expansion of $(a-b{)}^n$ is

• A. $\frac{(n-r+2)(n-r+1)}{r(r-1)}\cdot \frac{b^2}{a^2}$
• B. $\frac{(n-r+2)(n-r+1)}{r(r-1)}.\frac{a^2}{b^2}$
• C. $(\frac{n-r+2}{r})\cdot \frac{b}{a}$
• D. $(\frac{n-r+1}{r-1})\cdot \frac{b}{a}$

Answer 1

The correct option is A $\frac{(n-r+2)(n-r+1)}{r(r-1)}\cdot \frac{b^2}{a^2}$

We know, $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Therefore $\frac{T_{r+1}}{T_{r-1}}$
$=\frac{{}^n{C}_r{a}^{n-r}{b}^r}{{}^n{C}_{r-2}{a}^{n-r+2}{b}^{r-2}}$
$=\frac{{}^n{C}_r{b}^2}{{}^n{C}_{r-2}{a}^2}$
$=\frac{n!(n-(r-2\left)\right)!(r-2)!b^2}{n!(n-r)!r!a^2}$
$=\frac{(n-(r-2\left)\right)(n-(r-1\left)\right)b^2}{r(r-1)a^2}$
$=\frac{(n-r+2)(n-r+1)b^2}{r(r-1)a^2}$