A) 2
B) \[10/3\]
C) \[{{2}^{1000}}+1\]
D) \[\text{10}\]
Correct Answer: B
Solution :
\[\frac{{{2}^{2001}}+{{2}^{1999}}}{{{2}^{2000}}-{{2}^{1998}}}=\frac{{{2}^{1999}}({{2}^{2}}+1)}{{{2}^{1998}}({{2}^{2}}-1)}\] \[={{2}^{1999-1998}}\left( \frac{4+1}{4-1} \right)\] \[\left[ \because \frac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \right]\] \[=2\left( \frac{5}{3} \right)=\frac{10}{3}\]You need to login to perform this action.
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