Question

Suppose that $F(n+1)=\frac{2F\left(n\right)+1}{2}$ for $n=1,2,\mathrm{3....},$ and $F\left(1\right)=2$. Then $F\left(101\right)$ equals

• A. $50$
• B. $52$
• C. $54$
• D. None of these

Answer 1

Answer: (B)

$52$

Given:

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

$F\left(1\right)=2$

Putting $n=1$

$\therefore F\left(2\right)=\frac{2F\left(1\right)+1}{2}=2.5$

Putting $n=2$

$\therefore F\left(3\right)=\frac{2F\left(2\right)+1}{2}=3$

Putting $n=3$

$\therefore F\left(4\right)=\frac{2F\left(3\right)+1}{2}=3.5$

$\therefore$ Series we get:

$2,2.5,3,3.5,.....$

$\therefore a=2,d=0.5$

$\therefore F\left(101\right)=2+(101-1)0.5=52$