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(1^(2))/(1!)+(1^(2)+2^(2))/(2!)+(1^(2)+2...

`(1^(2))/(1!)+(1^(2)+2^(2))/(2!)+(1^(2)+2^(2)+3^(2))/(3!)+....=`

A

`17e//6`

B

`6e//17`

C

`11e//7`

D

`7e//11`

Text Solution

Verified by Experts

The correct Answer is:
A
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(1^(2) )/( 1) + (1^(2) + 2^(2) )/(1+2) + (1^(2) + 2^(2) + 3^(2) )/( 1+ 2+ 3)+ …. + n terms =

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(1^(2).2)/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(3!)+....=

The sum (3)/(1^(2)) + (5)/(1^(2) + 2^(2)) + (7)/(1^(2)+ 2^(2) + 3^(2)) + ….. upto 11 terms is

(1)/(1!)+(1+2)/(2!)+(1+2+2^(2))/(3!)+....=

(1)/(1!)+(1+3)/(2!)+(1+3+3^(2))/(3!)+....=

If 1^(2)+(2^(2))/(2!)+(3^(2))/(3!)+(4^(2))/(4!)+....=ae,(1^(2).2)/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(3!)+...=be,(1)/(2!)+(1+2)/(3!)+(1+2+3)/(4!)+...=ce then the descending order of a,b,c is