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Using Summation Notation

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How to Sum the First n Terms of an Arithmetic Sequence

How to Sum the First n Terms of a Geometric Sequence

Summation Notation is used to find the sum of some or all terms of a mathematical sequence. In this notation the symbol Σ, a stylized Sigma Greek letter, represents a summation math operation.

 n

ai

i = 1

This tells us that we are to sum the first n terms of the sequence (a1 + a2 + a3 + … + an). We also know to begin at the first term because the index i = 1; i is a variable subscript, the index of summation.

As an example we evaluate:


 6

(i – 1) / i 2 = 0/1 + 1/4 + 2/9 + 3/16 + 4/25 + 5/36 = 3451/3600

i = 1


The index of summation can begin at any integer value:


 4

k3 = 23 + 33 + 43 = 8 + 27 + 64 = 99

k = 2

Since the first n terms of an Arithmetic Sequence can be written as:


a1, a1 + d, a1 = 2d, a1 + 3d, … , a1 + (n – 1)d

(Where d is the Common Difference as defined on the Sequence and Series page)


Or,

a1, a1 + d, a1 + 2d, … , an – 2d, an – d, an


Then the sum of n terms can be found as follows:


sn = a1 + (a1 + d) + (a1 + 2d) + … + (an – 2d) + (an – d) + an


And, writing the sum backward:


sn = an + (an – d) + (an – 2d) + … + (a1 + 2d + (a1 + d) + a1


Adding both sn term by term we get:


2sn = (a1 + an) + (a1 + an) + (a1 + an) + (a1 + an) + (a1 + an) + (a1 + an)


Giving us 2sn = (n) (a1 + an) because there are n terms.


If 2sn = (n) (a1 + an)


Then sn must equal,


sn = (n) (a1 + an) / 2 where a1 is the first term and an is the nth term.

As an example we use the sequence set {5, 9, 13, 17}:


s = 05 + 09 + 13 + 17 = 44

s = 17 + 13 + 09 + 05 = 44

      22 + 22 + 22 + 22


2s = 22 x 4 = 88

s = 88 / 2 = 44


s4 = (n) (a1 + an) / 2 = 4 (05 + 17) / 2 = 44


It is easy to see that if we want to sum a large sequence of data the Arithmetic Sum Formula is quicker, much more efficient.

The first n terms of a Geometric Sequence can be written as:


a1, a1 r, a1 r2, … , a1 r n – 1

(Where r is the Common Ratio as defined on the Sequence and Series page)


So the sum of the first n terms is:

sn = a1 + a1 r + a1 r2 + … + a1 r n – 1


Then,


r sn = a1 r2 + a1 r3 + … + a1 r n


So,


sn – r sn = a1 + (a1 r – a1 r) + (a1 r2 – a1 r2) + … + a1 r n – 1 – a1 r n – 1) – a1 r n


Thus, (by factoring, sn – r sn, and adding the terms to the right of the equals)


sn (1 – r) = a1 – a1 r n


sn = a1 (1 – r n) / (1 – r)


As an example we evaluate the following geometric series with a first term of 5 and common ratio of -3:


s6 = a1 (1 – r n) / (1 – r) = 5 (1 – (-36 ) ) / (1 – (-3)) = 3650 / 4 = 912 1/2

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