Find (a) the coefficient of x^7in theepansion of(ax^2+1/(bx))^11 (b)The coefficient of x^(-7) in the expansion of (ax^2+1/(bx))^11 Also , find the relation between a and b, so that these coefficients are equal .

Find (a) the coefficient of x^7in theepansion of(ax^2+1/(bx))^11 (b)T

1.	Find (a) the coefficient of x^7in theepansion of(ax^2+1/(bx))^11 (b)The coefficient of x^(-7) in the expansion of (ax^2+1/(bx))^11 Also , find the relation between a and b, so that these coefficients are equal .
Answer» Solution :In the expansion of `(ax^2+(1)/(BX))^11`, the general term is :, `T_(r+1)=^(11)C_(r)(ax^2)^(11-r)(1/(bx))^r=11C_(r).(a^(11-r))/(b^r).x^(22-3r)` PUTTING 22-3r = 7 `therefore3r= 15rArr r= 5 ` `therefore T_(6)=^(11)C_(5)(a^6)/(b^(5)).X^7` Hence the cofficient of `x^7 in (ax^2+(1)/(bx))^11is ""^11 C_(5)a^6 b^(-5)` NOTE that BINOMIAL coefficient of sixth term is `""^(11)C_(5)` In the expansion of `(ax^2+(1)/(bx^2)^11` general term is `T_(r+1)=""^11C_r(ax)^(11-r)((-1)/(bx^2))^r=(-1)^(r 11) C_(r)(a^(11-r))/(b^(r)).X^(11-3r)` putting 11-3r = -7 `therefore T_(6)=^(11)C_(5)(a^6)/(b^(5)).X^7` Hence the coffieceint of `x^7 "in" (ax^2+(1)/(bx))^11 is ""^(11)C_(6)a^5b^(-6)` Also given : Coefficient of `x^7 "in" (ax -(1)/(bx^2))^11=" coeffiecient of " x-7 "in" (ax-(1)/(bx^2))^11` `rArr""C_(5)a^(6)b^(-5)=""11C_(6)a^5b^(-6)` `rArrab=1 ""(therefore""^C_(5)=""^(11)C_(6))` which is the required relation between a and b .