16/(2!)+(64)/(3!) + (256)/(4!) + ..... o...

`16/(2!)+(64)/(3!) + (256)/(4!) + ..... oo` =

A

`e^(4) -1 `

B

`e^(4) -2 `

C

`e^(4) -4 `

D

`e^(4) -5 `

Text Solution

Verified by Experts

The correct Answer is:

D

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AAKASH SERIES-EXPONENTIAL SERIES -EXERCISE - III

16/(2!)+(64)/(3!) + (256)/(4!) + ..... oo =
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(1+x^2/(2!)+x^4/(4!)+....oo)^2 =
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1 +(2^3)/(1!) x + (3^3)/(2!) x^2 + ....oo =
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If (e^x)/(1-x)=B0 +B1 x +B2 x^2 + ....... + B(n-1) x^(n-1) +B(n)x^n+ ....
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If C =1 + (cos x ) /(1!) +(cos 2X)/(2!) + (cos3x)/(3!) + ......and
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Coefficient of x^n in e^(e^x) is 1/(n!) k then k =
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Assertion (A) : (3^2)/(2!) +(3^4)/(4!)+(3^6)/(4!)+ .... = Cosh 3 -1 ...
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