Question

If $f(n+1)=\frac{2f\left(n\right)+1}{2},n=1,2,.....$ and $f\left(1\right)=2$, then $f\left(101\right)=..........$

• A. $53$
• B. $52$
• C. $51$
• D. $50$

Answer 1

Answer: (B)

$52$

Given:
$\Rightarrow f(n+1)=\frac{2f\left(n\right)+1}{2}$
and $\Rightarrow f\left(1\right)=2$
For $n=1$,$f\left(2\right)=\frac{2f\left(1\right)+1}{2}=\frac{5}{2}$

For $n=2$, $f\left(3\right)=\frac{2f\left(2\right)+1}{2}=3$

For $n=3$,$f\left(4\right)=\frac{2f\left(3\right)+1}{2}=\frac{6+1}{2}=\frac{7}{2}$

So, $\Rightarrow f\left(1\right),f\left(2\right),f\left(3\right),f\left(4\right),....=2,\frac{5}{2},3,\frac{7}{2},.....$

$\therefore f\left(n\right)=\frac{3+n}{2}$

$\Rightarrow f\left(101\right)=\frac{3+101}{2}=52$

$\Rightarrow f\left(101\right)=52$
Hence, option 'B' is correct.