Solveeit Logo
Login

Question

Mathematics Question on binomial expansion formula

The value of the term independent of x in the expansion of (1+x2−2x)4,x≠0\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \ne 0 is equal to

A

1

B

-6

C

-5

D

6

Answer

-5

Explanation

Solution

(1+x2−2x)4\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4} =4C0+4C1(x2−2x)+4C2(x2−2x)2= ^{4}C_{0} + ^{4}C_{1} \left(\frac{x}{2}-\frac{2}{x}\right)+^{4}C_{2} \left(\frac{x}{2}-\frac{2}{x}\right)^{2} 4C3(x2−2x)3+4C4(x2−2x)4=4C0+4C1(x2−2x)+4C2[x24−2+4x2]^{4}C_{3} \left(\frac{x}{2}-\frac{2}{x}\right)^{3}+^{4}C_{4} \left(\frac{x}{2}-\frac{2}{x}\right)^{4} = ^{4}C_{0} + ^{4}C_{1}\left(\frac{x}{2}-\frac{2}{x}\right)+ ^{4}C_{2} \left[\frac{x^{2}}{4}-2+\frac{4}{x^{2}}\right] +4C3[3C0(x2)3−3C1(x2)2(2x)+3C2(x2)(2x)2−3C3(2x)3]+ ^{4}C_{3} \left[ ^{3}C_{0}\left(\frac{x}{2}\right)^{3}- ^{3}C_{1} \left(\frac{x}{2}\right)^{2} \left(\frac{2}{x}\right)+ ^{3}C_{2} \left(\frac{x}{2}\right) \left(\frac{2}{x}\right)^{2}- ^{3}C_{3} \left(\frac{2}{x}\right)^{3}\right] +4C0[4C0(x2)4−4C1(x2)3(2x)+4C2(x2)2(2x)2−4C3(x2)(2x)3+4C4(2x)4]+ ^{4}C_{0} \left[ ^{4}C_{0}\left(\frac{x}{2}\right)^{4}- ^{4}C_{1} \left(\frac{x}{2}\right)^{3} \left(\frac{2}{x}\right)+ ^{4}C_{2} \left(\frac{x}{2}\right)^{2} \left(\frac{2}{x}\right)^{2}- ^{4}C_{3} \left(\frac{x}{2}\right) \left(\frac{2}{x}\right)^{3} + ^{4}C_{4}\left(\frac{2}{x}\right)^{4}\right] The term independent of x in above =4C0+4C2(−2)+4C4.4C2=1−12+6=−5= ^{4}C_{0} +^{4}C_{2}\left(-2\right) + ^{4}C_{4}. ^{4}C_{2} = 1-12 + 6 = -5