A) \[{{16.2}^{22}}\]
B) \[8.\text{ }{{2}^{20}}\]
C) \[{{8.2}^{21}}\]
D) \[{{16.2}^{21}}\]
Correct Answer: D
Solution :
\[{{2.}^{20}}{{C}_{0}}+{{5.}^{20}}{{C}_{1}}+{{8.}^{20}}{{C}_{2}}+...+{{62.}^{20}}{{C}_{20}}\] |
\[=\sum\limits_{r=0}^{20}{(3r+2){{.}^{20}}{{C}_{r}}}\] |
\[=3\sum\limits_{r=0}^{20}{{{r}_{0}}{{.}^{20}}{{C}_{r}}}+2\sum\limits_{r=0}^{20}{^{20}{{C}_{r}}}\] |
\[=3\times 20\sum\limits_{r=0}^{20}{^{19}{{C}_{r-1}}}+2.({{2}^{20}})\] |
\[={{60.2}^{19}}+{{2.2}^{20}}\] |
\[={{16.2}^{21}}.\] |
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