Sums & Series - Math

Sums & Series

Suppose

is a sequence.

a1, a2, ...

Sometimes we'll want to sum the first numbers (also known as

)

k

terms

that appear in a sequence. A shorter way to write + + + ? ? ? + is as

a1 a2 a3

ak

Xk

ai

i=1

P

There are four rules that are important to know when using . They are

listed below. In all of the rules,

and

are sequences

a1, a2, a3, ... b1, b2, b3, ...

and c 2 R.

Xk

Xk

Rule 1

=

. c ai

cai

i=1

i=1

Rule #1 is the distributive law. It's another way of writing the equation

( + +???+ )= + +???+

c a1 a2

ak ca1 ca2

cak

Xk

Xk

Xk

Rule 2 .

+

= (+)

ai

bi

ai bi

i=1

i=1

i=1

This rule is essentially another form of the commutative law for addition. It's another way of writing that

( + +???+ )+( + +???+ ) = ( + )+( + )+???+( + )

a1 a2

ak b1 b2

bk a1 b1 a2 b2

ak bk

Xk

Xk

Xk

Rule 3

.

ai

=(

)

bi

ai bi

i=1

i=1

i=1

25

Rule #3 is a combination of the first two rules. To see that, remember that

= ( 1) , so we can use Rule #1 (with = 1) followed by Rule #2 to

bi

bi

c

derive Rule #3, as is shown below:

Xk ai

i=1

Xk

Xk

Xk

=

+

bi

ai

bi

i=1

i=1

i=1

Xk = ( + ( ))

ai bi

i=1

Xk

=(

)

ai bi

i=1

Xk

Rule 4.

c = kc

i=1

The fourth rule can bPe a little tricky. The number c does not depend on i

-- it's a constant -- so k is taken to mean that you should add the first

c

i=1

terms in the sequence

. That is to say that

k

c, c, c, c, ...

Xk c = c + c + ? ? ? + c = kc

i=1

Examples. P

? 5 2 means that you should add the first 5 terms of the constant

i=1

sequence 2 2 2 2 2 . That is, , , , , ,...

X5 2 = 2 + 2 + 2 + 2 + 2 = 5(2) = 10

i=1

P ? 20 3 = 20(3) = 60

i=1

*************

26

Sum of first terms in an arithmetic sequence k

If

a1,

a2,

a3,

...

is

an

arithmetic

sequence,

then

an+1

=

an

+ d

for

some

d

2

R.

We want to show that

Xk = k( + )

ai 2 a1 an

i=1

To show this, let's write the sum in question in two dierent ways: front-to-

back, and back-to-front. That is,

Xk =

ai a1

+( + )+( +2 )+???+( 2 )+(

)+

a1 d a1 d

ak d ak d ak

i=1

and Xk = ai ak

+(

)+( 2 )+???+( +2 )+( + )+

ak d ak d

a1 d a1 d a1

i=1

Add the two equations above "top-to-bottom" to get

Xk

2

=[ + ]+[ + ]+[ + ]+???+[ + ]+[ + ]+[ + ]

ai a1 ak a1 ak a1 ak

a1 ak a1 ak a1 ak

i=1

Count and check that there are exactly of the [ + ] terms in the line

k

a1 ak

above being added. Thus,

Xk

2

=[ + ]

ai k a1 ak

i=1

which is equivalent to what we were trying to show:

Xk = k( + )

ai 2 a1 ak

i=1

Example. What is the sum of the first 63 terms of the sequence 1 2 5 8 ? , , , , ...

The sequence above is arithmetic, because each term in the sequence is 3 plus the term before it, so = 3. The first term of the sequence is 1, so

d

27

= 1. Our formula = +( 1) tells us that = 1+(62)3 = 185.

a1

an a1 n d

a63

Therefore,

X 63

63

63

= ( 1 + 185) = (184) = 5 796

ai 2

2

,

i=1

Example. The sum of the first 201 terms of the sequence 10, 17, 24, 31, ...

equals 201(10 + 1410) = 201(1420) = 142 710. ,

2

2

*************

Geometric series

It usually doesn't make any sense at all to talk about adding infinitely

many numbers. But if

is a geometric sequence where =

a1, a2, a3, ...

an+1 ran

and 1

1, then we can make sense of adding all of the terms of

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