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The sum of the series 1+1/(4*2!)+1/(16...

The sum of the series ` 1+1/(4*2!)+1/(16*4!)+1/(64*6!)+...` is

A

` (e+1)/(2sqrt(e))`

B

` (e-1)/(2sqrt(e))`

C

` (e+1)/(sqrt(e))`

D

` (e-1)/(sqrt(e))`

Text Solution

Verified by Experts

The correct Answer is:
a
We know that , ` (e^(x)+e^(-x))/2 = 1+ (x^(2))/(2!) + (x^(4))/(4!) + (x^(6))/(6!) +...`
On putting `x = 1/2 ,` we get
` (e^(1//2) +e^(-1//2))/2 = 1+(1/2)^(2) 1/(2!) + (1/2)^(4) 1/(4!) +...`
` rArr (e+1)/(2sqrt(2)) = 1+ 1/(4*2!) +1/(16*4!)+ ...oo`
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Knowledge Check

  • The sum of the series 1 + (1)/(64.4!) + (1)/(64.6!) + ...oo is

    A
    `(e-1)/(2sqrt(e))`
    B
    `(e+1)/(2sqrt(e))`
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    `(e-1)/(sqrte)`
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    `e^-`

  • The sum of the series (1)/(2!) - (1)/(3!) + (1)/(4!) + ... is

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    e
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    `e^(-1//2)`
    C
    `e^(-2)`
    D
    `e^(-1)`

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    A
    `(e+1)/(2sqrte)`
    B
    `(e-1)/(2sqrte)`
    C
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