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JEE Mains · Maths · STD 11 - 7. binomial theoram

Let \(\alpha=\sum_{\mathrm{r}=0}^{\mathrm{n}}\left(4 \mathrm{r}^2+2 \mathrm{r}+1\right)^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}\) and \(\beta=\left(\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1}\right)+\frac{1}{\mathrm{n}+1}\). If \(140<\frac{2 \alpha}{\beta}<281\) then the value of \(n\) is ...........

  1. A \(9\)
  2. B \(4\)
  3. C \(5\)
  4. D \(6\)
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Answer & Solution

Correct Answer

(C) \(5\)

Step-by-step Solution

Detailed explanation

\( \alpha=\sum_{\mathrm{r}=0}^{\mathrm{n}}\left(4 \mathrm{r}^2+2 \mathrm{r}+1\right) \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \)…
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