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2‐2 Series
Unit 2 Sequences and Series
Concepts and Objectivesp j
Series (Obj. #7)( j )Given an arithmetic or geometric series, be able to calculate Sn, the nth partial sum and vice versaId tif h th t i i d itIdentify whether a geometric series converges and its limit
Series
A series is the indicated sum of the terms of a sequence.qSince a sequence has an infinite number of terms, a series can be thought of as a sum of an infinite numbers f t Th f ll f th t ill th fof terms. The sum of all of the terms will, therefore, usually be infinite.For this reason, it is convenient to study sums of only aFor this reason, it is convenient to study sums of only a finite number of terms of a series. The sum of part of a series is called a partial sum.
Series
The sum of the first n terms of a series is called the nthpartial sum of that series. It is usually represented by Sn.Example: For the sequence 3, 5, 7, 9, …, find S4.
Series
The sum of the first n terms of a series is called the nthpartial sum of that series. It is usually represented by Sn.Example: For the sequence 3, 5, 7, 9, …, find S4.
=1 3S= + =2 3 5 8S
+ +3 5 7 15S
We could also write this as
= + + =3 3 5 7 15S= + + + =4 3 5 7 9 24S
We could also write this as
=
= + =∑4
412 1 24
k
S k (“sigma” notation)
Series
An arithmetic or geometric series is a series which gresults from adding the terms of an arithmetic or geometric sequence, respectively.Whil it i ibl t fi d ti l b iti dWhile it is possible to find a partial sum by writing down all of the terms and adding them up, it would easier to find a formula for Sn.
Series
Consider the arithmetic sequence 7, 13, 19, 25, 31, …q , , , , ,What if we wanted to find the sum of the first 100 terms?We can find the 100th term using the formula from before: Since d = 6 we can work backwards from 601 to find the
( )( )= + − =100 7 100 1 6 601tSince d = 6, we can work backwards from 601 to find the last few terms as well, so our sum looks like:
= + + + + + + + +100 7 13 19 25 ... 583 589 595 601S100
Series
Now, what do you notice if we add the terms together , y gstarting from the outside and working inside?
= + + + + + + + +100 7 13 19 25 ... 583 589 595 601S100
Series
Now, what do you notice if we add the terms together , y gstarting from the outside and working inside?
= + + + + + + + +100 7 13 19 25 ... 583 589 595 601S100
608
608608
608
608We have 50 sums (100 ÷ 2), so our sum is 50(608)=30400. This leads us to our formula.
608
Series
The formula for the nth partial sum of an arithmetic pseries is
( ) +⎛ ⎞= + = ⎜ ⎟⎝ ⎠
11 or
2 2n
n n n
t tnS t t S n
Since if we don’t know tn, we can find it by the sequence formula we can substitute that into the above formula
⎝ ⎠2 2
formula, we can substitute that into the above formula and get
( )( )= + + −1 1 1nS t t n d( )( )+ +1 1 1
2nS t t n d
( )( )= + −12 12n
t n d2
Series
Example: Find S33 for the series with t1 = 9 and d = –2.p 33 1
Series
Example: Find S33 for the series with t1 = 9 and d = –2.p 33 1
( )( )= + −12 12n
nS t n d
( ) ( )( )( )= + −33 2 9 32 22
Notice that even though n is odd we still get a whole
( )= − = −33 46 7592
Notice that even though n is odd, we still get a whole number. Why does the formula work this way?
Series
You may recall our first geometric sequence, which was y g q ,3, 6, 12, 24, 48, …
This series has t1 = 3 and r = 2, and the terms looked like
1 3t =
2 3 2t = i2
3 3 2 2 3 2t = =i i i3
4 3 2 2 2 3 2t = =i i i i13 2nt −= i3 2nt = i
Series
If we wanted the sum of the first 100 terms of this sequence, we could write it as follows:
= + + + + + +i i i i i2 3 98 99100 3 3 2 3 2 3 2 ... 3 2 3 2S
Now, suppose we multiplied both sides of this equation by –2 or the opposite of our common ratio and addedby –2, or the opposite of our common ratio, and added the two equations together:
= + + + + + +i i i i i2 3 98 99100 3 3 2 3 2 3 2 ... 3 2 3 2S
− = − − − − − − −i i i i i i i2 3 98 99 1001002 3 2 3 2 3 2 ... 3 2 3 2 3 2S
Series
= + + + + + +i i i i i2 3 98 993 3 2 3 2 3 2 3 2 3 2S = + + + + + +
− = − − − − − − −
i i i i i
i i i i i i i100
2 3 98 99 100100
3 3 2 3 2 3 2 ... 3 2 3 2
2 3 2 3 2 3 2 ... 3 2 3 2 3 2
S
S
1002 3 0 0 0 0 0 3 2S S− = + + + + + + − i 100100 1002 3 0 0 0 ... 0 0 3 2S S
− = − i 100100 1002 3 3 2S S
Solving for S, gives us( ) ( )− = −
⎛ ⎞
100100
100
1 2 3 1 2
1 2
S
which certainly suggests a formula.
⎛ ⎞−= ⎜ ⎟−⎝ ⎠
100
1001 231 2
S
which certainly suggests a formula.
Series
The nth partial sum of a geometric series is given by the p g g yformula
⎛ ⎞−= ⎜ ⎟1
1 nrS t
F I’ l t t d t t t f t th
⎜ ⎟−⎝ ⎠1 1nS t
r
For some reason, I’m always tempted to try to factor the fraction further. It doesn’t factor.
Series
Example: Find S34 for the geometric series with t1 = 7 p 34 g 1and r = 1.03.
Series
Example: Find S34 for the geometric series with t1 = 7 p 34 g 1and r = 1.03.
Using the formula, we have:⎛ ⎞−
= ⎜ ⎟⎝ ⎠
34
341 1.0371 1 03
S ⎜ ⎟−⎝ ⎠34 1 1.03≈ 404.111
Series
Example: 30702 is a partial sum in the arithmetic series p pwith first term 17 and common difference 3. Which partial sum is it? (What is n?)
Series
Example: 30277 is a partial sum in the arithmetic series p pwith first term 17 and common difference 3. Which partial sum is it? (What is n?)
( )( )= + −12 12n
nS t n d
( )( )( )30277 2 17 1 3n ( )( )( )= + −i30277 2 17 1 32
n
( )= +30277 31 3nn( )
2= + 260554 31 3n n
Series
(cont.) + − =23 31 60554 0n n( )− + − =23 411 442 60554 0n n n
( ) ( )− + − =3 137 442 137 0n n n( ) ( )+3 137 442 137 0n n n
( )( )+ − =3 442 137 0n n
442
So n = 137
= −442 , 1373
n
So n = 137.
Series
Example: 50238.14 is the approximate value of a partial p pp psum in the geometric series with t1 = 150 and r = 1.04. Which term is it?
Series
Example: 50238.14 is the approximate value of a partial p pp psum in the geometric series with t1 = 150 and r = 1.04. Which term is it?
⎛ ⎞1 nr⎛ ⎞−= ⎜ ⎟−⎝ ⎠
111n
rS t
r
⎛ ⎞−1 1 04n⎛ ⎞−= ⎜ ⎟−⎝ ⎠
1 1.0450238.14 1501 1.04
( )( )−50238.14 0.04( )( )= −
50238.14 0.041 1.04
150n
=1.04 14.39683...n
Series
(cont.) Taking the log of each side gets n out of the ( ) g g gexponent.
=log1.04 log14.39683...n
=log1.04 log14.39683...n
= =log14.39683... 68.000001...
l 1 04n
So n = 68.log1.04