Question:

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

Updated On: Jul 7, 2022
  • 1
  • -6
  • -5
  • 6
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The Correct Option is C

Solution and Explanation

$\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}$ $= ^{4}C_{0} + ^{4}C_{1} \left(\frac{x}{2}-\frac{2}{x}\right)+^{4}C_{2} \left(\frac{x}{2}-\frac{2}{x}\right)^{2}$ $^{4}C_{3} \left(\frac{x}{2}-\frac{2}{x}\right)^{3}+^{4}C_{4} \left(\frac{x}{2}-\frac{2}{x}\right)^{4} = ^{4}C_{0} + ^{4}C_{1}\left(\frac{x}{2}-\frac{2}{x}\right)+ ^{4}C_{2} \left[\frac{x^{2}}{4}-2+\frac{4}{x^{2}}\right]$ $+ ^{4}C_{3} \left[ ^{3}C_{0}\left(\frac{x}{2}\right)^{3}- ^{3}C_{1} \left(\frac{x}{2}\right)^{2} \left(\frac{2}{x}\right)+ ^{3}C_{2} \left(\frac{x}{2}\right) \left(\frac{2}{x}\right)^{2}- ^{3}C_{3} \left(\frac{2}{x}\right)^{3}\right]$ $+ ^{4}C_{0} \left[ ^{4}C_{0}\left(\frac{x}{2}\right)^{4}- ^{4}C_{1} \left(\frac{x}{2}\right)^{3} \left(\frac{2}{x}\right)+ ^{4}C_{2} \left(\frac{x}{2}\right)^{2} \left(\frac{2}{x}\right)^{2}- ^{4}C_{3} \left(\frac{x}{2}\right) \left(\frac{2}{x}\right)^{3} + ^{4}C_{4}\left(\frac{2}{x}\right)^{4}\right]$ The term independent of x in above $= ^{4}C_{0} +^{4}C_{2}\left(-2\right) + ^{4}C_{4}. ^{4}C_{2} = 1-12 + 6 = -5 $
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Concepts Used:

Binomial Expansion Formula

The binomial expansion formula involves binomial coefficients which are of the form 

(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.

This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:

We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn

General Term = Tr+1 = nCr xn-r . yr

  • General Term in (1 + x)n is nCr xr
  • In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .