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
- A \(3\)
- B \(4\)
- C \(5\)
- D \(6\)
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…
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