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1+(2)/(3!)+(3)/(5!)+(4)/(7!)+....=...

`1+(2)/(3!)+(3)/(5!)+(4)/(7!)+....=`

A

e

B

`e//2`

C

`e^(-1)`

D

`e^(-1)//2`

Text Solution

Verified by Experts

The correct Answer is:
B
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I : (2)/(1!)+(4)/(3!)+(6)/(5!)+....=e-1 II : (3)/(2!)+(5)/(4!)+(7)/(6!)+(9)/(8!)+....=e-1

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(1)/(1.3)+(1)/(2)((1)/(3.5))+(1)/(3)((1)/(5.7))+....=

If A=[((2)/(3), 1,(5)/(3)),((1)/(3), (2)/(3), (4)/(3)),((7)/(3), 2, (2)/(3))] and B=[((2)/(5), (3)/(5), 1),((1)/(5), (2)/(5), (4)/(5)),((7)/(5),(6)/(5),(2)/(5))] , then compute 3A-5B .

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