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`(2)/(2!)+(2+4)/(3!)+(2+4+6)/(4!)+….infty` is equal to
A. `e-2`
B. `e-1`
C. e
D. `e^(-1)`

1 Answer

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Best answer
Answer:
Let `T_(n)` be the `n^(th)` term of the given series then
`T_(n)=(2+4+6+..+2n)/(n+1)!,n=1,2,3………`
`rarr T_(n)=(n(n+1))/(n+1)!=(1)/(n-1)!,n=1,2…`
`therefore "Sum of the series" =underset(n=1)overset(infty)Sigma T_(n)=underset(n=1)overset(infty)Sigma (1)/(n-1)!=e`

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