Question

Find the co-efficient of $x^7$ in ${(a{x}^2+\frac{1}{bx})}^{11}$ and $x^{-7}$ in $x^7$ in ${(ax-\frac{1}{b{x}^2})}^{11}$ and find the relation between 'a' & 'b' so that these co-efficients are equal. (where a, $b\ne 0)$

Answer 1

$(a{x}^2+\frac{1}{bx})$

$T_{r+1}={}^{11}{C}_r(a{x}^2{)}^{11-r}.{\left(\frac{1}{bx}\right)}^r$

$={}^{11}{C}_r{a}^{11-r}.x^{22-2r-r}.{\left(\frac{1}{b}\right)}^r$

$={}^{11}{C}_r.a^{11-r}.\frac{1}{b^r}.x^{22-3r}$

$22-3r=7$

$3r=15$

$r=5$

$T_6={}^{11}{C}_r.\frac{a^6}{b^5}.x^5$

${(ax-\frac{1}{b{x}^2})}^{11}$

$T_{r+1}={}^{11}{C}_r(ax{)}^{11-r}{\left(\frac{-1}{b{x}^2}\right)}^r$

$={}^{11}{C}_r\frac{a^{11-r}}{(-b{)}^r}.x^{11-r-2r}$

$={}^{11}{C}_r.\frac{a^{11-r}}{(-b{)}^r}.x^{11-3r}$

$11-3r=-7$

$3r=18$

$r=6$

$T_7={}^{11}{C}_6\frac{a^5}{b^6}.x^{-7}$

Coefficient are equal

$\frac{a^5}{b^6}=\frac{a^6}{b^5}$

$ab=1$