AP 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
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
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,…
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