(3)/(2!)+(5)/(4!)+(7)/(6!)+(9)/(8!)+.......

`(3)/(2!)+(5)/(4!)+(7)/(6!)+(9)/(8!)+....=`

A

e

B

`2e`

C

`e^(-1)`

D

`e-1`

Text Solution

Verified by Experts

The correct Answer is:

D

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DIPTI PUBLICATION ( AP EAMET)-EXPONENTIAL SERIES & LOGARITHMIC SERIES (APPENDIX-1)-EXERCISE 1A

1+(1)/(3!)+(1)/(5!)+....=
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(2)/(1!)+(4)/(3!)+(6)/(5!)+....=
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(3)/(2!)+(5)/(4!)+(7)/(6!)+(9)/(8!)+....=
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(2)/(3!)+(4)/(5!)+(6)/(7!)+....=
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(1)/(3!)+(2)/(5!)+(3)/(7!)+....=
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1+(3)/(1!)+(5)/(2!)+(7)/(3!)+....=
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(3)/(2!)+(5)/(3!)+(7)/(4!)+(9)/(5!)+....=
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1+(4)/(2!)+(7)/(3!)+(10)/(4!)+....=
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underset(n=1)overset(oo)Sigma(2n)/((2n+1)!)=
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underset(n=1)overset(oo)sum(n^(2))/((n+1)!)=
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underset(n=1)overset(oo)Sigma(2n^(2)+n+1)/(n!)=
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