Crossover-101(BITSAT-2015)

Calculus Level 5

Suppose A=n=01(2n)! \large \displaystyle A=\sum _{ n=0 }^{ \infty }{ \frac { 1 }{ (2n)! } } .

Then which one of the following is equal to A2A^2?

  1. 12+n=122n1(2n)!\large \displaystyle \frac{1}{2}+\sum _{ n=1 }^{ \infty }{ \frac { { 2 }^{ 2n-1 } }{ (2n)! } }

  2. n=01((2n)!)2\large \displaystyle \sum _{ n=0}^{ \infty }{ \frac { 1 }{ { ((2n)!) }^{ 2 } } }

  3. 1+n=122n1(2n)!\large \displaystyle 1+\sum _{ n=1 }^{ \infty }{ \frac { { 2 }^{ 2n-1 } }{ (2n)! } }

  4. 12+n=122n(2n)!\large \displaystyle \frac{1}{2}+\sum _{ n=1 }^{ \infty }{ \frac { { 2 }^{ 2n} }{ (2n)! } }

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