Consider the sequence of natural numbers `s_0,s_1,s_2`,... such that `s_0 =3, s_1 = 3 and s_n = 3 + s_(n-1) s_(n-2)`, then

Options

A. `(e^(pi))/(2)`
B. `e^(-pi)`
C. `e^(pi)`
D. `(e^(-pi))/(2)`

Answer

C

Explanation

`"Since "-1 le sin x le 1," the curve "y=e^(-x) sin x " is bounded by the curve "y=e^(-x)and y=e^(-x).`

Also, the curve `y=e^(-x) sin x` intersects the positive semi-axis OX at the points where sin x=0, where `x_(n)=npi, n in Z.`
`"Also "|y_(n)|=|y" coordinate in the half-wave "S_(n)|`
`therefore" "S_(n)=(-1)^(n)overset((n+1)pi)underset(npi)inte^(-x)sin xdx`
`=((-1)^(n+1))/(2)[e^(-x)(-sin x + cos x)]_(npi)^((n+1)pi)`
`=((-1)^(n+1))/(2)[e^(-(n+1)x)(-1)^(n+1)-e^(npi)(-1)^(n)]`
`=(e^(-npi))/(2)(1+e^(pi))`
`rArr" "(S_(n+1))/(S_(n))=e^(-pi)`
`"and "S_(0)=(1)/(2)(1+e^(pi)).`
Therefore, the sequence `S_(0),S_(1),S_(2),...` forms an infinite G.P. with common ratio `e^(-pi)`. we have
`overset(oo)underset(n=0)SigmaS_(n)=((1)/(2)(1+e^(pi)))/(1-e^(-pi))`