Answer to Question #231602 in Discrete Mathematics for shavneet

Question

find 200 sum k=100^k

Answer 1

$$\displaystyle\sum_{k=100}^{200}k=\displaystyle\sum_{k=1}^{200}k-\displaystyle\sum_{k=1}^{99}k$$

$$=\dfrac{200(200+1)}{2}-\dfrac{99(99+1)}{2}=$$

$$=50(402-99)=15150$$