A) 400
B) 50
C) 200
D) 100
Correct Answer: D
Solution :
| \[\sum\limits_{i=1}^{20}{{{\left( \frac{^{20}{{C}_{I-1}}}{^{20}{{C}_{I}}{{+}^{20}}{{C}_{I-1}}} \right)}^{3}}}\] Now \[\left( \frac{{}^{20}{{C}_{I-1}}}{{}^{20}{{C}_{I-1}}+{}^{20}{{C}_{I-1}}} \right)=\frac{{}^{20}{{C}_{I-1}}}{{}^{21}{{C}_{I}}}=\frac{I}{21}\] |
| Let given sum be S, so |
| \[S=\sum\limits_{I=1}^{20}{\frac{{{(i)}^{3}}}{{{21}^{3}}}=\frac{1}{{{(21)}^{3}}}{{\left( \frac{20.21}{2} \right)}^{2}}}=\frac{100}{21}\] |
| Given \[S=\frac{k}{21}\]\[\Rightarrow \]k =100. |
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