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If S(n)=(1.2)/(3!)+(2.2^(2))/(4!)+(3.2^(...

If `S_(n)=(1.2)/(3!)+(2.2^(2))/(4!)+(3.2^(2))/(5!)+...+` up to `n` terms, then sum of infinite terms is

A

`(4)/(pi)`

B

`(3)/(e)`

C

`(pi)/(r )`

D

`1`

Text Solution

Verified by Experts

The correct Answer is:
D
`(d)` Here `t_(r )=(r*2^(r ))/((r+2)!)=((r+2-2)2^(r ))/((r+2)!)=(2^(r ))/((r+1)!)-(2^(r+1))/((r+2)!)`
`:.S_(n)=sum_(r=1)^(n)t_(r )=sum_(r=1)^(n)(2^(r ))/((r+1)!)-(2^(r+1))/((r+2)!)=1-(2^(n+1))/((n+2)!)`
`:.S_(oo)=lim_(ntooo)(1-(2^(n+1))/((n+2)!))=1`
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