If A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} and B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} , then {A \over B} is equal to :

Question

Accepted Answer

Answer: C. -{{11} \over 9} A = \sum\limits_{n = 1}^\infty {{1 \over {{{(3 + {{( - 1)}^n})}^n}}}} and B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{(3 + {{( - 1)}^n})}^n}}}} A = {1 \over 2} + {1 \over {{4^2}}} + {1 \over {{2^3}}} + {1 \over {{4^4}}} + ........ B = {{ - 1} \over 2} + {1 \over {{4^2}}} - {1 \over {{2^3}}} + {1 \over {{4^4}}} + ...... A = {{{1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}}, B = {{ - {1 \over 2}} \over {1 - {1 \over 4}}} + {{{1 \over {16}}} \over {1 - {1 \over {16}}}} A = {{11} \over {15}}, B = {{ - 9} \over {15}} 	herefore {A \over B} = {{ - 11} \over 9}

If $$A = \sum\limits_{n = 1}^\infty {{1 \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$ and $$B = \sum\limits_{n = 1}^\infty {{{{{( - 1)}^n}} \over {{{\left( {3 + {{( - 1)}^n}} \right)}^n}}}} $$, then $${A \over B}$$ is equal to :

Answer

Explanation

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