Calculate the sum of 1001 terms of the series 3,5,7,...

We notice that if we'll calculate the difference between 2
consecutive terms of the given series, we'll obtain the same value each
time:

5 - 3 = 7 - 5 = ...... =
2

So, the given series is an arithmetic progression whose
common difference is d = 2.

We can calculate the sum of n
terms of an arithmetic progression in this way;

Sn = (a1 +
an)*n/2

a1 - the first term of the
progression

a1 = 3

an - the
n-th term of the progression

an =
a1001

a1001 = a1 +
(1001-1)*d

a1001 = 3 +
1000*2

a1001 = 3 + 2000

a1001
= 2003

n = 1001 - the number of
terms

 S1001 = [a1 + a1 +
(1001-1)*d]*1001/2

 S1001 = (3 +
2003)*1001/2

 S1001 =
2006*1001/2

 S1001 =
1003*1001

The sum of the 1001 terms of the
series 3,5,7,... is:

 S1001 =
1004003