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
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When you see this think sets
This is a consecutive number set from 1 to 99 so the number of terms is 99
the mean (avg) of the terms is (1+99)/2 = 50
so the sum = mean x number of terms = 50 x 99 = 4950
This is a consecutive number set from 1 to 99 so the number of terms is 99
the mean (avg) of the terms is (1+99)/2 = 50
so the sum = mean x number of terms = 50 x 99 = 4950
you could also use this formula
[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)
[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)