If F(n+1)=(2F(n)+1)/(2) for n=1, 2, …, and F(1)=2, then F(101) equals:

1966 AMC 12 Problem 25

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25.

F(n+1)=\dfrac{2F(n)+1}{2}

for

n=1,

2,

\ldots,

and

F(1)=2,

then

F(101)

equals:

49

50

51

52

53

Simplify the recurrence to

F(n+1)=F(n)+\frac{1}{2}

Count the increments from

n=1

n=101

Each step adds

\frac{1}{2}.

There are

100

steps from

F(1)

F(101),

F(101)=2+\frac{100}{2}=52.

Therefore, the correct answer is D.