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

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

A

27e

B

`6e-1`

C

7e

D

2e

Text Solution

Verified by Experts

The correct Answer is:
A
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(1^(2).2)/(1!)+(2^(2).3)/(2!)+(3^(2).4)/(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

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

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

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

(1.2)/(1!)+(2.3)/(2!)+(3.4)/(3!)+....=

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

Find the determinant of the matrix [(1^(2),2^(2),3^(2)),(2^(2),3^(2),4^(2)),(3^(2),4^(2),5^(2))]