A) \[{{2}^{24}}\]
B) \[{{2}^{25}}\]
C) \[{{2}^{26}}\]
D) \[{{2}^{23}}\]
Correct Answer: B
Solution :
\[{{2.}^{20}}{{C}_{0}}+{{5.}^{20}}{{C}_{1}}+{{8.}^{20}}{{C}_{2}}+{{11.}^{20}}{{C}_{3}}+\]\[...+{{62.}^{20}}{{C}_{20}}\] \[\sum\limits_{r=0}^{20}{{{(3r+2)}^{20}}{{C}_{r}}}\] \[=3\sum\limits_{r=0}^{20}{r{{.}^{20}}{{C}_{r}}}+2\sum\limits_{r=0}^{20}{^{20}{{C}_{r}}}\] \[=3\sum\limits_{r=0}^{20}{r}\left( \frac{20}{r} \right){{\,}^{19}}{{C}_{r-1}}+{{2.2}^{20}}\] \[={{60.2}^{19}}+{{2.2}^{20}}={{2}^{25}}\]
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