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TS EAMCET · Maths · Binomial Theorem

If \(p\) is an integral multiple of 4 lying in between the coefficients of \(x^4\) and \(x\) in the expansion of \(\left(x^2+\frac{1}{x}\right)^8\), then the number of such values of \(p\) is

  1. A \(3\)
  2. B \(4\)
  3. C \(5\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(3\)

Step-by-step Solution

Detailed explanation

The general term in the expansion of \(\left(x^2+\frac{1}{x}\right)^8\) is \(T_{r+1}={ }^8 C_r\left(x^2\right)^{8-r}\left(\frac{1}{x}\right)^r={ }^8 C_r x^{16-3 r}\) For the coefficient of \(x^4\), put \(r=4\), we get…