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

*print*Print*list*Cite

### 2 Answers

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.

Adding 1001 terms yields:

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

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