SATHEE: Permutations and Combinations 2 Question 7

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

(d) 50

Show Answer

Answer:

Correct Answer: 7. (a)

Solution:

Given,

$\begin{aligned} \sum_{i = 1}^{20} \frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}} = \frac{k}{21} \\ \Rightarrow & \sum_{i = 1}^{20} \frac{^{20} C_{i - 1}}{^{21} C_{i}} = \frac{k}{21} (∵^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}) \\ \Rightarrow & \sum_{i = 1}^{20} \frac{^{20} C_{i - 1}}{\frac{21}{i}^{20} C_{i - 1}} = \frac{k}{21} ∵^{n} C_{r} = \frac{n}{r}^{n - 1} C_{r - 1} \\ \Rightarrow & \sum_{i = 1}^{20} \frac{i}{21} = \frac{k}{21} \\ \Rightarrow & \frac{1}{(21)^{3}} \sum_{i = 1}^{20} i^{3} = \frac{k}{21} \end{aligned}$

$\begin{matrix} \Rightarrow \frac{1}{(21)^{3}} \frac{n (n + 1)}{2}_{n = 20}^{2} = \frac{k}{21} \\ ∵ 1^{3} + 2^{3} + \dots + n^{3} = \frac{n (n + 1)^{2}}{2} \\ \Rightarrow k = \frac{21}{(21)^{3}} \frac{20 \times 21^{2}}{2} = 100 \\ k = 100 \end{matrix}$

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

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Permutations and Combinations 2 Question 7

7. If ∑i=12020Ci−120Ci+20Ci−1=k21, then k equals

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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7. If $\sum_{i = 1}^{20} \frac{^{20} C_{i - 1}}{^{20} C_{i} +^{20} C_{i - 1}} = \frac{k}{21}$ , then $k$ equals