The coefficient of x6 in the power series expansion of (x4-12 x2+7/(x2+1)3) is

Question

The coefficient of x6 in the power series expansion of (x4-12 x2+7/(x2+1)3) is (A) 149 (B) -253 (C) -145 (D) 253

Tardigrade Admin · Accepted Answer

We have, (x4-12 x2+7/(x2+1)3)=(x4-12 x2+7)(x2+1)-3 =(x4-12 x2+7)(1-3 x2+(12 x4/2 !)-(60/3 !) x6 ldots) =(x4-12 x2+7)(1-3 x2+6 x4-10 x6 ldots .) Coefficient of x6 is (-3-72-70)=-145

Q. The coefficient of x6 in the power series expansion of (x2+1)3x4−12x2+7​ is

149

-253

-145

253

We have, (x2+1)3x4−12x2+7​=(x4−12x2+7)(x2+1)−3 =(x4−12x2+7)(1−3x2+2!12x4​−3!60​x6…) =(x4−12x2+7)(1−3x2+6x4−10x6….) Coefficient of x6 is (−3−72−70)=−145

Q. The coefficient of $x^{6}$ in the power series expansion of $\frac{x ^{4} - 12 x ^{2} + 7}{( x ^{2} + 1 ) ^{3}}$ is