Permutations and Combinations 2 Question 7
7. If $\sum _{i=1}^{20} \frac{{ }^{20} C _{i-1}}{{ }^{20} C _i+{ }^{20} C _{i-1}}=\frac{k}{21}$, then $k$ equals
(a) 100
(b) 400
(c) 200
(d) 50
Show Answer
Answer:
Correct Answer: 7. (a)
Solution:
- Given,
$$ \begin{aligned} & \sum _{i=1}^{20} \frac{{ }^{20} C _{i-1}}{{ }^{20} C _i+{ }^{20} C _{i-1}}=\frac{k}{21} \\ \Rightarrow \quad & \sum _{i=1}^{20} \frac{{ }^{20} C _{i-1}}{{ }^{21} C _i}=\frac{k}{21} \quad\left(\because{ }^{n} C _r+{ }^{n} C _{r-1}={ }^{n+1} C _r\right) \\ \Rightarrow \quad & \sum _{i=1}^{20} \frac{{ }^{20} C _{i-1}}{\frac{21}{i}{ }^{20} C _{i-1}}=\frac{k}{21} \quad \because{ }^{n} C _r=\frac{n}{r}{ }^{n-1} C _{r-1} \\ \Rightarrow \quad & \quad \sum _{i=1}^{20} \frac{i}{21}=\frac{k}{21} \\ \Rightarrow \quad & \frac{1}{(21)^{3}} \sum _{i=1}^{20} i^{3}=\frac{k}{21} \end{aligned} $$
$$ \begin{gathered} \Rightarrow \quad \frac{1}{(21)^{3}} \frac{n(n+1)}{2}{ } _{n=20}^{2}=\frac{k}{21} \\ \because 1^{3}+2^{3}+\ldots+n^{3}=\frac{n(n+1)^{2}}{2} \\ \Rightarrow \quad k=\frac{21}{(21)^{3}} \frac{20 \times 21^{2}}{2}=100 \\ k=100 \end{gathered} $$
