JEE Math Practice Question
The product $${2^{{1 \over 4}}}{.4^{{1 \over {16}}}}{.8^{{1 \over {48}}}}{.16^{{1 \over {128}}}}$$ ... to $$\infty $$ is equal to :
- A.$${2^{{1 \over 4}}}$$
- B.$${2^{{1 \over 2}}}$$
- C.1
- D.2
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