JEE Mains · Maths · STD 11 - 7. binomial theoram
If the constant term, in binomial expansion of \(\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}\) is \(180,\) than \(r\) is equal to \(......\)
- A \(1\)
- B \(2\)
- C \(6\)
- D \(8\)
Answer & Solution
Correct Answer
(D) \(8\)
Step-by-step Solution
Detailed explanation
\(\left(2 x^{r}+\frac{1}{x^{2}}\right)^{10}\) \(\text { General term }={ }^{10} C_{R}\left(2 x^{r}\right)^{10-R} x^{-2 R}\) \(\Rightarrow 2^{10-R 10} C_{R}=180 \ldots \ldots . .(1)\) \(\,(10-R) r-2 R=0\) \(r=\frac{2 R}{10-R}\) \(r=\frac{2(R-10)}{10-R}+\frac{20}{10-R}\)…
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