Solve this following

Question:

$\sum_{\mathrm{i}=1}^{20}\left(\frac{{ }^{20} \mathrm{C}_{\mathrm{i}-1}}{{ }^{20} \mathrm{C}_{\mathrm{i}}+{ }^{20} \mathrm{C}_{\mathrm{i}-1}}\right)=\frac{\mathrm{k}}{21}$, then $\mathrm{k}$ equals :

 

  1. 200

  2. 50

  3. 100

  4. 400


Correct Option: , 3

Solution:

$\sum_{i=1}^{20}\left(\frac{{ }^{20} \mathrm{C}_{\mathrm{i}-1}}{{ }^{20} \mathrm{C}_{\mathrm{i}}+{ }^{20} \mathrm{C}_{\mathrm{i}-1}}\right)^{3}=\frac{\mathrm{k}}{21}$

$\Rightarrow \sum_{\mathrm{i}=1}^{20}\left(\frac{{ }^{20} \mathrm{C}_{\mathrm{i}-1}}{{ }^{21} \mathrm{C}_{\mathrm{i}}}\right)^{3}=\frac{\mathrm{k}}{21}$

$\Rightarrow \sum_{\mathrm{i}=1}^{20}\left(\frac{\mathrm{i}}{21}\right)^{3}=\frac{\mathrm{k}}{21}$

$\Rightarrow \frac{1}{(21)^{3}}\left[\frac{20(21)}{2}\right]^{2}=\frac{\mathrm{k}}{21}$

$\Rightarrow 100=\mathrm{k}$

 

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