ધારો કે alpha=sum_{r=0}^nleft(4 r^2+2 r | vedclass

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Question

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

Vedclass · Accepted Answer

$5$

Vedclass · Answer

$ \alpha=\sum_{\mathrm{r}=0}^{\mathrm{n}}\left(4 \mathrm{r}^2+2 \mathrm{r}+1ight) \cdot{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} $

$ \alpha=4 \sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r}^2 \cdot \frac{\mathrm{n}}{\mathrm{r}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}+2 \sum_{\mathrm{r}=0}^{\mathrm{n}} \mathrm{r} \cdot \frac{\mathrm{n}}{\mathrm{r}} \cdot{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-}+\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} $

$ +4 n \sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}+2 \mathrm{n} \sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}+\sum_{\mathrm{r}=0}^{\mathrm{n}}{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} $

$ \alpha=4 \mathrm{n}(\mathrm{n}-1) \cdot 2^{\mathrm{n}-2}+4 \mathrm{n} \cdot 2^{\mathrm{n}-1}+2 \mathrm{n} \cdot 2^{\mathrm{n}-1}+2^{\mathrm{n}} $

$ \alpha=2^{\mathrm{n}-2}[4 \mathrm{n}(\mathrm{n}-1)+8 \mathrm{n}+4 \mathrm{n}+4] $

$ \alpha=2^{\mathrm{n}-2}\left[4 \mathrm{n}^2+8 \mathrm{n}+4ight] $

$ \alpha=2 \mathrm{n}(\mathrm{n}+1)^2 $

$ \beta=\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1}+\frac{1}{\mathrm{n}+1} $

$ =\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{r}+1}}{\mathrm{n}+1}+\frac{1}{\mathrm{n}+1} $

$ =\frac{1}{\mathrm{n}+1}\left(1+{ }^{\mathrm{n}+1} \mathrm{C}_1+\ldots .+{ }^{\mathrm{n}+1} \mathrm{C}_{\mathrm{n}+1}ight) $

$ =\frac{2^{\mathrm{n}+1}}{\mathrm{n}+1} $

$ \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n+1)^2}{2^{n+1}} \cdot(n+1)=(n+1)^3 $

$ 140<(\mathrm{n}+1)^3<281 $

$ \mathrm{n}=4 \Rightarrow(\mathrm{n}+1)^3=125 $

$ \mathrm{n}=5 \Rightarrow(\mathrm{n}+1)^3=216 $

$ \mathrm{n}=6 \Rightarrow(\mathrm{n}+1)^3=343 $

$ 	herefore \mathrm{n}=5 $

ધારો કે $\alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right)^n C_r$ અને $\beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1} \cdot$ જો $140 < \frac{2 \alpha}{\beta}<281$ તો $n$ નું મૂલ્ય .......... છે.

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