# Difference between revisions of "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 $2+50=\boxed{\text{(D)} \ 52}.$

Solution by davidaops.