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 ______ .

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Binomial Theorem

Let

α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1)^{n} C_{r}

and

β = (\sum_{r = 0}^{n} \frac{^{n} C _{r}}{r + 1}) + \frac{1}{n + 1}

. If

140 < \frac{2 α}{β} < 281

, then the value of

n

is ______ .

Answer: 5

α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1)^{n} C_{r} = 2 n (n + 1) 2^{n - 2} \cdot 2 = 2 n (n + 1)^{2} \cdot 2^{n - 1}

(after evaluating the standard sums), and

β = \sum_{r = 0}^{n} \frac{^{n} C _{r}}{r + 1} + \frac{1}{n + 1} = \frac{2 ^{n + 1}}{n + 1}

. Then

\frac{2 α}{β} = (n + 1)^{3}

. From

140 < (n + 1)^{3} < 281

(n + 1)^{3} = 216

gives

n + 1 = 6

, so

n = 5

JEE Main · 2024 · 08 Apr 2024 · Shift 1 · Integer answer