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7.Binomial Theorem
hard
$\sum_{ i =1}^{20}\left(\frac{{ }^{20} C _{ i -1}}{{ }^{20} C _{ i }+{ }^{20} C _{ i -1}}\right)^{3}=\frac{ k }{21}$, तो $k$ बराबर है
A
$400$
B
$50$
C
$200$
D
$100$
(JEE MAIN-2019)
Solution
${\sum\limits_{i = 1}^{20} {\left( {\frac{{{\,^{20}}{C_{I – 1}}}}{{^{20}{C_1} + {\,^{20}}{C_{I – 1}}}}} \right)} ^3}$
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
$S = \sum\limits_{I = 1}^{20} {\frac{{{{\left( i \right)}^3}}}{{{{21}^3}}}} = \frac{1}{{{{(21)}^3}}}{\left( {\frac{{20.21}}{2}} \right)^2} = \frac{{100}}{{21}}$
Given $S = \frac{k}{{21}} \Rightarrow k = 100$
Standard 11
Mathematics
