Question

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

• A. 50
• B. 52
• C. 54
• D. none of these

Answer 1

The correct option is B

52

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

$\Rightarrow F(n+1)-F\left(n\right)=1/2$

Hence, the given series is an A.P. with common difference 1/2 and first term being 2. F (101) is 101${}^{st}$ term of A.P. given by 2+(101-1) (1/2)=52.