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Question: In a P.D., the sum of the terms in the odd places exceeds the sum of the terms in the even places by...

In a P.D., the sum of the terms in the odd places exceeds the sum of the terms in the even places by 14\frac { 1 } { 4 } , then P(X=2)

A

12(log⁡2)2\frac { 1 } { 2 } ( \log 2 ) ^ { 2 }

B

16(log⁡2)2\frac { 1 } { 6 } ( \log 2 ) ^ { 2 }

C

14(log⁡2)2\frac { 1 } { 4 } ( \log 2 ) ^ { 2 }

D

18(log⁡2)2\frac { 1 } { 8 } ( \log 2 ) ^ { 2 }

Answer

14(log⁡2)2\frac { 1 } { 4 } ( \log 2 ) ^ { 2 }

Explanation

Solution

eλ(cosh⁡λ−sinh⁡λ)=14\mathrm { e } ^ { \lambda } ( \cosh \lambda - \sinh \lambda ) = \frac { 1 } { 4 }

e−λ(eλ+e−λ2−eλ−e−λ2)=14\mathrm { e } ^ { - \lambda } \left( \frac { \mathrm { e } ^ { \lambda } + \mathrm { e } ^ { - \lambda } } { 2 } - \frac { \mathrm { e } ^ { \lambda } - \mathrm { e } ^ { - \lambda } } { 2 } \right) = \frac { 1 } { 4 } ; e−λ=12,λ=log⁡2\mathrm { e } ^ { - \lambda } = \frac { 1 } { 2 } , \lambda = \log 2

P(X=2) =