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Find (a) the coefficient of `x^7 ` in the epansion 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 .

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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
`therefore 3r= 15 rArr 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))^11 is ""^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)`
`rArr ab=1 " "(therefore ""^C_(5)=""^(11)C_(6))`
which is the required relation between a and b .

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