Whats the formula for these types of problems?

1+2+3+4+...+99=

A 4700

B 4750

C 4850

D 4900

E 4950

## num props

##### This topic has expert replies

[n(n+1)]/2

1+2+3+4+...+99

so n=99

(99*100)/2 =4950

this is a relatively easy problem if you know the formula. also be aware that this formula only works for consecutive positive integers and when the sequence starts from 1 to n.

PS (this should be a 500-600 level math problem IMO)