# SumFind the sum of 1001 terms of the series 3,5,7,...

sciencesolve | Certified Educator

Notice the relation between any two successive terms of series. Each term generates the next term if 2 is added to it. This relation proves that the sequence is arithmetic progression. The common difference is the constant ammount added to the previous term to get the next term.

3 + 5 + 7 + ....+a1001 = 1001*(3+a1001)*1001/2

You need to calculate the 1001st term:

a1001=3+2*1000 => a1001= 2003

3 + 5 + 7 + ....+a1001 = 1001*(3+2003)/2 => 3 + 5 + 7 + ....+a1001 = 1001*2006/2 => 3 + 5 + 7 + ....+a1001 = 2008006/2

3 + 5 + 7 + ....+a1001 = 1004003

giorgiana1976 | Student

Computing the difference between 2 consecutive terms of the given series, we'll get the same value:

5 - 3 = 7 - 5 = ...... = 2

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

The identity that gives the sum of n terms of an arithmetic progression is:

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