JEE Main & Advanced Mathematics Sequence & Series Question Bank Self Evaluation Test - Sequences and Series

If \[{{S}_{n}}=\]\[(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}}),\] then \[{{S}_{\infty }}\] is equal to

A) 1

B) \[\frac{1}{2}\]

C) \[\frac{3}{2}\]

D) None

Correct Answer: C

Solution :

[c] \[{{S}_{n}}=(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-4}})....(1+{{3}^{-2n}})\]

\[\Rightarrow (1-{{3}^{-1}}){{S}_{n}}\]

\[=(1-{{3}^{-1}})(1+{{3}^{-1}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}})\]

\[\Rightarrow \frac{2}{3}{{S}_{n}}=(1-{{3}^{-2}})(1+{{3}^{-2}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-2n}})\]

\[=(1-{{3}^{-4}})(1+{{3}^{-4}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}})\]

\[=(1-{{3}^{-8}})(1+{{3}^{-8}})....(1+{{3}^{-{{2}^{n}}}})\]

\[=(1-{{3}^{-{{2}^{n}}}})(1+{{3}^{-{{2}^{n}}}})=1-{{({{3}^{-{{2}^{n}}}})}^{2}}=1-{{3}^{-{{2}^{n+1}}}}\]

\[\Rightarrow {{S}_{n}}=\frac{3}{2}(1-{{3}^{-{{2}^{n+1}}}})\]

\[\therefore {{S}_{\infty }}=\frac{3}{2}(1-0)=\frac{3}{2}\]