Permutations And Combinations Ques 35
- If $\sum _{i=1}^{20} \frac{{ }^{20} C _{i-1}}{{ }^{20} C _i+{ }^{20} C _{i-1}}=\frac{k}{21}$, then $k$ equals
100
400
200
50
Show Answer
Answer:
Correct Answer: 35.(a)
Solution:
Formula:
- Given,
$ \begin{aligned} & \sum _{i=1}^{20} \left(\frac{{ }^{20} C _{i-1}}{{ }^{20} C _i+{ }^{20} C _{i-1}}\right)^{3} =\frac{k}{21} \\ \Rightarrow \quad & \sum _{i=1}^{20} (\frac{{ }^{20} C _{i-1}}{{ }^{21} C _i}) ^ {3}=\frac{k}{21} \quad\left(\because{ }^{n} C _{r}+{ }^{n} C _{r-1}={ }^{n+1} C _{r}\right) \\ \Rightarrow \quad & \sum _{i=1}^{20} (\frac{{ }^{20} C _{i-1}}{\frac{21}{i}{ }^{20} C _{i-1}}) ^ {3} =\frac{k}{21} \quad \because{ }^{n} C _r=\frac{n}{r}{ }^{n-1} C _{r-1} \\ \Rightarrow \quad & \quad \sum _{i=1}^{20} (\frac{i}{21}) ^ {3}=\frac{k}{21} \\ \Rightarrow \quad & \frac{1}{(21)^{3}} \sum _{i=1}^{20} i^{3}=\frac{k}{21} \end{aligned} $
$ \begin{gathered} \Rightarrow \quad \frac{1}{(21)^{3}} \frac{n(n+1)}{2} \sum_{n=2}^{20}=\frac{k}{21} \\ \because 1^{3}+2^{3}+\ldots+n^{3}=\left(\frac{n(n+1)}{2}\right)^{2} \ \Rightarrow \quad k=\frac{21}{(21)^{3}} \frac{20 \times 21^{2}}{2}=100 \\ k=100 \end{gathered} $
BETA
