Question

If $f(n+1)=f\left(n\right)+n$ for all $n\ge 0$ or $f\left(0\right)=1$, then $f\left(200\right)$ equals

• A. $21100$
• B. $21000$
• C. $20900$
• D. $19901$

Answer 1

Answer: (D)

$19901$

We have,

$\begin{array}{c}f\left(n+1\right)=f\left(n\right)+n\\ f\left(200\right)=\frac{n}{2}\left(a+l\right)+1\\ =\frac{199}{2}\left(1+199\right)+1\\ f\left(200\right)=19901\end{array}$

Hence, this is the answer.