underset(n=1)overset(oo)sum(n^(2))/((n+1...

`underset(n=1)overset(oo)sum(n^(2))/((n+1)!)=`

A

e

B

2e

C

3e

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+(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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(1+(1)/(2!)+(1)/(4!)+(1)/(6!)+....)^(2)-(1+(1)/(3!)+(1)/(5!)+(1)/(7!)+...
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(1+(a^(2)x^(2))/(2!)+(a^(4)x^(4))/(4!)+....)^(2)-(ax+(a^(3)x^(3))/(3!)...
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(1+(2^(2))/(2!)+(2^(4))/(3!)+(2^(6))/(4!)+....)/(1+(1)/(2!)+(2)/(3!)+(...
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(1+(1)/(3!)+(1)/(5!)+....)/(1+(1)/(2!)+(1)/(4!)+....)=
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((1)/(2!)+(1)/(4!)+(1)/(6!)+....)/(1+(1)/(3!)+(1)/(5!)+....)=
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(1)/(2)-(1)/(3(1!))+(1)/(4(2!))-(1)/(5(3!))+....=
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(1)/(1!)+(1+2)/(2!)+(1+2+2^(2))/(3!)+....=
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(1)/(1!)+(1+3)/(2!)+(1+3+3^(2))/(3!)+....=
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(1)/(2!)+(1+2)/(3!)+(1+2+3)/(4!)+....
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(2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+....=
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(1)/(1!)+(1+3)/(2!)+(1+3+3^(2))/(3!)+....=
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1^(2)+(2^(2))/(2!)+(3^(2))/(3!)+(4^(2))/(4!)+....=
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(1^(2))/(2!)+(2^(2))/(3!)+(3^(2))/(4!)+....=
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(1^(2))/(1!)+(1^(2)+2^(2))/(2!)+(1^(2)+2^(2)+3^(2))/(3!)+....=
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(1^(2).2)/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(3!)+....=
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(1^(2).2^(2))/(1!)+(2^(2).3^(2))/(2!)+(3^(2).4^(2))/(3!)+....=
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