# Thread: maths - sum of a series

1. ## maths - sum of a series

a. find the sum of the first 200 pos. intergers of
1 + 2 + 3 + 4 + . . . + 200
is it
Tn =a+(n-1)d
200=1+(n-1)1
=1+n-1
200=n
n=200

then
Sn=n/2(a+l)
S200=200/2(1+200)
=100 X 201
=20100
is that right?

b. series 1 + 5 + 7 + 11 + . . . + 199 is formed by omitting all those which are multiples of 2 or 3 up to the first 200 intergers. Find the sum

how do you do this??

2. Your first sum is correct. Think 100 pairs of numbers with each pair adding up to 201 (1+200, 2+199, 3+198, etc)

The 2nd sum can be done in the same way. Add up all the numbers from 1 to 199. Subtract the series:
2+4+6+8+...198
Subtract the series
3+6+9+12+...198
Add back the series
6+12+18+...198.
I think this is called inclusion/exclusion.

3. ok so you find the sum of the whole lot
then minus the sum of the 2 multiples
then minus the sum of the 3 multiples
then add back the sum of the other