1966 AHSME Problems/Problem 25

Problem

If $F(n+1)=\frac{2F(n)+1}{2}$ for $n=1,2,\cdots$ and $F(1)=2$ , then $F(101)$ equals:

$\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53$

Solution

Notice that $\frac{2F(n)+1}{2}=F(n)+\frac{1}{2}.$

This means that for every single increment $n$ goes up from $1$ , $F(n)$ will increase by $\frac{1}{2}.$ Since $101$ is $100$ increments from $1$ , $F(n)$ will increase $\frac{1}{2}\times100=50.$

Since $F(1)=2,$ $F(101)$ will equal $2+50=\boxed{52 \text{(D)}}.$

1966 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 24	Followed by Problem 26
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1966 AHSME Problems/Problem 25

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