(1/(2!)+1/(4!)+1/(6!)+ ....oo)/(1+1/(3!)...

`(1/(2!)+1/(4!)+1/(6!)+ ....oo)/(1+1/(3!)+1/(5!)+....oo) `

`(e+1)/(e-1)`

`(e)/(e-1)`

`1/(e-1)`

`(e-1)/(e+1)`

Verified by Experts

The correct Answer is:

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sum(n=1)^(oo) ((logex)^(2n-1))/((2n-1)!)=
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(1+1/(2!) + 1/(4!)+ .....)^2 - (1+1/(3!) + 1/(5!)+ .....)^2 =
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(1/(2!)+1/(4!)+1/(6!)+ ....oo)/(1+1/(3!)+1/(5!)+....oo)
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Match the following Lists. {:("List - I ","List - II "),((A)The (n+1...
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1/(3!) +2/(5!) + (3)/(7!) + .....oo =
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2/(2!) +5/(3!) + (7)/(4!) + (9)/(5!)+.....oo =
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1+3/(1!) +5/(2!) + (7)/(3!) + ......oo=
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If an = (1+2+3+......+n)/(n!) then a1 +a2+a3 + ....oo =
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Which of the following statement is true Statement - I : 1 +(1+2) ...
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Assertion (A) 1: (3)/(2!) +(5)/(4!) +(7)/(6!) + ..... = e Reason (R...
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1/(1!) +(1+3)/(2!) + (1+3+5)/(3!)+......oo =
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sum(r=1)^(oo) (3r^(2) +r +5))/(r!) =
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(1.4)/(0!) +(2.5)/(1!) + (3.6)/(2!)+...... =
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(1.2)/(1!) +(2.3)/(2!) + (3.4)/(3!)+......oo =
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(1^2.2)/(1!) +(2^2.3)/(2!) + (3^2.4)/(3!)+......oo =
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1+(1+5)/(2!) +(1+5+5^2)/(3!) + ......oo =
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Coefficient of x^(10) in e^(3x) is
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Coefficient of x^(10) in (1-2x)/e^x is
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Coefficient of x^(n) in (e^(5x)+e^(x))/(e^(2x)) is .....
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If n is odd natural number , then coefficient of x^n in (e^(5x)+e^x)/(...
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