JEE Mains · Maths · STD 11 - 7. binomial theoram
If \({\sum\limits_{i = 1}^{20} {\left( {\frac{{{}^{20}{C_{i - 1}}}}{{{}^{20}{C_i} + {}^{20}{C_{i - 1}}}}} \right)} ^3}\, = \frac{k}{{21}}\), then \(k\) equals
- A \(400\)
- B \(50\)
- C \(200\)
- D \(100\)
Answer & Solution
Correct Answer
(D) \(100\)
Step-by-step Solution
Detailed explanation
\({\sum\limits_{i = 1}^{20} {\left( {\frac{{{\,^{20}}{C_{I - 1}}}}{{^{20}{C_1} + {\,^{20}}{C_{I - 1}}}}} \right)} ^3}\) Now \(\frac{{^{20}{C_{I - 1}}}}{{^{20}{C_1} + {\,^{20}}{C_{I - 1}}}} = \frac{{^{20}{C_{I - 1}}}}{{^{20}{C_1}}} = \frac{1}{{21}}\) Let given sum be \(S\), so…
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