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TS EAMCET · Maths · Binomial Theorem

If the coefficient of \(x\) is in expansion of \(\left(x^2+\frac{k}{x}\right)^5\) is 270 , then \(k\) is equal to

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

Answer & Solution

Correct Answer

(C) \(3\)

Step-by-step Solution

Detailed explanation

General term in the expansion of \(\left(x^2+\frac{k}{x}\right)^5\) is \[ \begin{aligned} T_{r+1} & ={ }^5 C_r\left(x^2\right)^{5-r}\left(\frac{k}{x}\right)^r \\ & ={ }^5 C_r x^{10-3 r} \cdot k^r \end{aligned} \] Let this term contains \(x\) then,…