Barisan dan deret 1 bilingual

KONSEP BARISAN DAN DERET

THE CONCEPT OF SEQUENCE AND SERIES

Hal.: 3 BARISAN DAN DERET AdaptifHal.: 3

Pola Barisan dan Deret Bilangan

Kompetensi Dasar :

Menerapkan konsep barisan dan deret aritmatika

Indikator :1. Nilai suku ke- n suatu barisan aritmatika ditentukan

menggunakan rumus

2. Jumlah n suku suatu deret aritmatika ditentukan dengan

menggunakan rumus


The Pattern of Sequence and Series Number

Basic Competence:

Applying the concept of arithmetic sequence and

series

Indicator :1. The value of n-th term in an arithmetic sequence is defined

by formula

2. The sum of n in term of arithmetic sequence is defined by

formula


Saat mengendarai motor, pernahkah kalian mengamati speedometer pada motor tersebut?

Pada speedometer terdapat angka-angka 0,20, 40, 60, 80, 100, dan 120 yang menunjukkan kecepatan motor saat kalian

mengendarainya. Angka-angka ini berurutan mulai dariyang terkecil ke yang terbesar dengan pola tertentu sehinggamembentuk sebuah pola barisan



When you ride a motor cycle, have you ever look at the speeedometer?

In speedometer,there are numbers of 0,20, 40, 60, 80, 100, and 120 which show the speed of your motor cycle. These numbers are un order, starts from the smallest to the biggest with certain pattern, so that it forms a pattern of sequence



Bayangkan anda seorang penumpang taksi. Dia harus membayar biaya buka pintu Rp 15.000 dan argo Rp 2.500 /km.

15.000 17.500 20.000 22.500 …….

Buka pintu 1 km 2 km 3 km 4 km


Hal.: 8 BARISAN DAN DERET Adaptif

Imagine that you are a taxi passenger. You have to pay the starting fee Rp 15.000 and it charge Rp 2.500 /km.

15.000 17.500 20.000 22.500 …….

Starting fee 1 km 2 km 3 km 4 km



NOTASI SIGMA

Konsep Notasi Sigma

Perhatikan jumlah 6 bilangan ganjil pertama berikut: 1 + 3 + 5 + 7 + 9 + 11 ……….. (1)

Pada bentuk (1) Suku ke-1 = 1 = 2.1 – 1Suku ke-2 = 3 = 2.2 – 1Suku ke-3 = 5 = 2.3 – 1Suku ke-4 = 7 = 2.4 – 1Suku ke-5 = 9 = 2.5 – 1Suku ke-6 = 11 = 2.6 – 1

Secara umum suku ke-k pada (1) dapat dinyatakan dalam bentuk 2k – 1, k { 1, 2, 3, 4, 5, 6 }


SIGMA NOTATION

The Concept of Sigma Notation

Look at the sum of the first sixth odd number below: 1 + 3 + 5 + 7 + 9 + 11 ……….. (1)

In the form(1) The 1st term = 1 = 2.1 – 1The 2nd term= 3 = 2.2 – 1The 3rd term = 5 = 2.3 – 1The 4th term = 7 = 2.4 – 1The 5th term = 9 = 2.5 – 1The 6th term = 11 = 2.6 – 1

Generally, the k-th term in (1) can be stated in the form of 2k – 1, k { 1, 2, 3, 4, 5, 6 }


NOTASI SIGMA

Dengan notasi sigma bentuk penjumlahan (1) dapatditulis :

6

1k

1)-(2k1197531


SIGMA NOTATION

In Sigma notation, the addition form (1) can be written as:

6

1k

1)-(2k1197531


Bentuk

6

1)12(

kk

dibaca “sigma 2k – 1 dari k =1 sampai dengan 6”

atau “jumlah 2k – 1 untuk k = 1 sd k = 6”

1 disebut batas bawah dan

6 disebut batas atas,

k dinamakan indeks

(ada yang menyebut variabel)

9

4)1)3(2(

kk

9

4)72(

kk

NOTASI SIGMA


In the form of

6

1)12(

kk

It is read “sigma 2k – 1 from k =1 to 6” or “the sum

of 2k – 1 for k = 1 sd k = 6”

1 is called lower limit and

6 is called upper limit,

k is called index (some people

called it variable)

9

4)1)3(2(

kk

9

4)72(

kk

SIGMA NOTATION


NOTASI SIGMA

Secara umum


SIGMA NOTATION

Generally


Nyatakan dalam bentuk sigma

1. a + a2b + a3b2 + a4b3 + … + a10b9

10

1k)1kbk(a

)142()132()122()112()12(4

1

k

k

Contoh:

249753

Hitung nilai dari:

NOTASI SIGMA


Stated into sigma form

1. a + a2b + a3b2 + a4b3 + … + a10b9

10

1k)1kbk(a

)142()132()122()112()12(4

1

k

k

Example:

249753

Define the value of

SIGMA NOTATION


NOTASI SIGMA

nnn

1n bCabC...baCbaCbaCa n1n

33nn3

22nn2

1nn1

n

n

0r

rrnnr baC

2. (a + b)n =


SIGMA NOTATION

nnn

1n bCabC...baCbaCbaCa n1n

33nn3

22nn2

1nn1

n

n

0r

rrnnr baC

2. (a + b)n =


Sifat-sifat Notasi Sigma :

, Untuk setiap bilangan bulat a, b dan n

.....1 3211

n

n

k

aaaaak

n

mk

n

mk

akCCak.2

n

mk

n

mk

n

mk

bkakbkak )(.3

pn

pmk

n

mk

pakak.4

CmnCn

mk

)1(.5

n

mk

n

pk

p

mk

akakak1

.6

0.71

m

mk

ak

NOTASI SIGMA


The properties of sigma notation :

, For every integer a, b and n

.....1 3211

n

n

k

aaaaak

n

mk

n

mk

akCCak.2

n

mk

n

mk

n

mk

bkakbkak )(.3

pn

pmk

n

mk

pakak.4

CmnCn

mk

)1(.5

n

mk

n

pk

p

mk

akakak1

.6

0.71

m

mk

ak

SIGMA NOTATION


NOTASI SIGMA

Contoh1:

Tunjukkan bahwa

Jawab :

3

1

3

1

)24()24(jk

ji

30)33.4()22.4()21.4()24(3

1

i

i

30)23.4()22.4()21.4()24(3

1

j

j


SIGMA NOTATION

Example 1:

Show that

Answer :

3

1

3

1

)24()24(jk

ji

30)33.4()22.4()21.4()24(3

1

i

i

30)23.4()22.4()21.4()24(3

1

j

j


NOTASI SIGMA

6

4

23

1

2 66kk

kk

6

1

26

1

26

4

23

1

2 6666kkkk

kkkk

Hitung nilai dari

Contoh 2 :

Jawab:

= 6 (12 +22 + 32 + 42 + 52 + 62)

= 6 (1 + 4 + 9 + 16 + 25 + 36)

= 6.91 = 546


SIGMA NOTATION

6

4

23

1

2 66kk

kk

6

1

26

1

26

4

23

1

2 6666kkkk

kkkk

Define the value of

Example 2 :

Answer:

= 6 (12 +22 + 32 + 42 + 52 + 62)

= 6 (1 + 4 + 9 + 16 + 25 + 36)

= 6.91 = 546


BARISAN DAN DERET ARITMATIKA

Bilangan-bilangan berurutan seperti pada speedometer memiliki selisih yang sama untuk setiap dua suku berurutannya sehingga membentuk suatu barisan bilangan

Barisan Aritmatika adalah suatu barisan dengan selisih (beda) dua suku yang berurutan selalu tetap Bentuk Umum : U1, U2, U3, …., Un

a, a + b, a + 2b,…., a + (n-1)b

Pada barisan aritmatika,berlaku Un – Un-1 = b sehingga Un = Un-1 + b


ARITHMETIC SEQUENCE AND SERIES

The orderly numbers like in speedometer have the same difference for every two orderly term, so it forms a sequence

Arithmetic sequence is sequence with difference two orderly term constant

The general form is : U1, U2, U3, …., Un

a, a + b, a + 2b,…., a + (n-1)b

In arithmetic sequence, we have Un – Un-1 = b, so Un = Un-1 + b


If you start arithmetic sequence with the first term a and difference b, then you will get this following sequence

The n-th term of arithmetic sequence is Un = a + ( n – 1 )b

Where Un = n-th term

a = the first term

b = difference

n = the term’s quantity


a a + b a + 2b a + 3b …. a + (n-1)b

Hal.: 31 BARISAN DAN DERET AdaptifHl.: 31



If every term of arithmetic sequence is added, then we will get arithmetic series.

Arithmetic series is the sum of terms of arithmetic sequence

General form :

U1 + U2 + U3 + … + Un atau

a + (a +b) + (a+2b) +… + (a+(n-1)b)

The formula of the sum of the first term in arithmetic series is

Where S = the sum of n-th term

n = the quantity of term

a = the first term

b = difference

= n-th term


bnan

Sn )1(22


Known: the sequence of 5, -2, -9, -16,…., find:

a.The formula of n-th term

b.The 25th term

Answer:

The difference of two orderly terms in sequence 5,-2, -9,-16 ,…is constant, b= -7,

so that the sequence is an arithmetic sequence

a.The formula of the n-th term in arithmetic sequence is

Un = 5 + ( n – 1 ). -7

Un = 5 + - 7n + 7

Un = -7n + 12

b. The 25th term of arithmetic sequence is : U12 = - 7.12 + 12

= - 163



Barisan geometri adalah suatu barisan dengan pembanding (rasio) antara dua suku yang berurutan selalu tetap.

Ada selembar kertas biru, akan dipotong-potong menjadi dua bagian.

BARISAN DAN DERET GEOMETRI


Geometric sequence is a sequence which has the constant ratio between two orderly term

There is blue paper. It will cut into two pieces

GEOMETRIC SEQUENCE AND SERIES


Look at the paper part that form a sequence

Every two orderly terms of the sequence have the same ratio

It seems that the ratio of every two orderly terms in the sequence is always constant. The sequence like this is called geometric sequence and the comparison of every two orderly term is called ratio (r)

1 2 4

U1 U2 U3

2....12

3

1

2 n

n

U

U

U

U

U

U



Geometric sequence is a sequence which have constant ratio for two orderly term

General form: U1, U2, U3, …., Un atau

a, ar, ar2, …., arn-1

In geometric sequence

If you start the geometric sequence with the first term a and the ratio is r, then you get the following sequence


rU

U

n

n 1

1. nn UrsehinggaU


Suku ke-n barisan Geometri adalah :



The n-th term of geometric sequence is :


Start With the first term a

Multiply with ratio r

Write the multiplication result



Hubungan suku-suku barisan geometriSeperti dalam barisan Aritmatika hubungan antara suku yang satu dan suku yang lain dalam barisan geometri dapat dijelaskan sebagai berikut:

Ambil U12 sebagai contoh :

U12 = a.r11

U12 = a.r9.r2 = U10. r2

U12 = a.r8.r3 = U9. r3

U12 = a.r4.r7 = U5. r7

U12 = a.r3.r8 = U4.r8

Secara umum dapat dirumuskan bahwa :

Un = Uk. rn-k



The relation of terms in geometric sequenceLike in arithmetic sequence, the relation between terms in geometric sequence can be explained as follows:

Take U12 as example :

U12 = a.r11

U12 = a.r9.r2 = U10. r2

U12 = a.r8.r3 = U9. r3

U12 = a.r4.r7 = U5. r7

U12 = a.r3.r8 = U4.r8

Generally, it can be formulated

Un = Uk. rn-k



Geometric series is the sum of terms in geometric sequenceGeneral form

U1 + U2 + U3 + …. + Un

a + ar + ar2 + ….+ arn-1

The formula of the n sum of the first term in geometric series is

1,1

)1(

rr

raS

n

n



Known sequence 27, 9, 3, 1, …..find

a.The formula of the n-th term

b. The 8th term

Answer:The ratio of two orderly terms in sequence 27,9,3, 1, …is constant,

so that the sequence is a geometric sequence

a. The formula of the n-th term in geometric sequence is

3

1r

1

3

127

n

nU

113 )3.(3 n

13 3.3 n

n 43



b. The 8th term of geometric sequence is

848 3 U

43

81

1

nnU

43


Deret geometi tak hingga adalah deret geometri yang banyak suku-sukunya tak hingga.Jika deret geometri tak hingga dengan -1 < r < 1 , maka jumlah deret geometri tak hingga tersebut mempunyai limit jumlah (konvergen).

Untuk n = ∞ , rn mendekati 0

Sehingga S∞ =

Dengan S∞ = Jumlah deret geometri tak hingga a = Suku pertama r = rasioJika r < -1 atau r > 1 , maka deret geometri tak hingganya akan divergen, yaitu jumlah suku-sukunya tidak terbatas

Deret Geometri tak hingga

r

a

1

r

raSn

n

1

)1(



Infinite geometric series is a geometric series which has infinite terms.If infinite geometric series is -1 < r < 1 , then the sum of geometric series has sum limit (convergent).

For n = ∞ , rn is close to 0

So S∞ =

With S∞ = the sum of infinite geometric series a = the first term r = ratioIf r < -1 or r > 1 , then the infinite geometric series will be divergent, means the sum of terms is not limited

Infinite Geometric Series

r

a

1

r

raSn

n

1

)1(



1. Hitung jumlah deret geometri tak hingga : 18 + 6 + 2 + … . .

Contoh :

3

1

2

3

1

2 u

u

u

ur


27

32

18

31

1

18

1

r

as

Jawab :

a = 18 ;


1. Find the sum of infinite geometric series : 18 + 6 + 2 + … . .

Example :

3

1

2

3

1

2 u

u

u

ur


27

32

18

31

1

18

1

r

as

Answer :

a = 18 ;


2. Sebuah bola elastis dijatuhkan dari ketinggian 2m. Setiap kali memantul dari lantai, bola mencapai ketinggian ¾ dari ketinggian sebelumnya. Berapakah panjang lintasan yang dilalui bola hingga berhenti ?


Lihat gambar di samping!Bola dijatuhkan dari A, maka AB dilalui satu kali, selanjutnya CD, EF dan seterusnya dilalui dua kali. Lintasannya membentuk deret geometri dengan a = 3 dan r = ¾ Panjang lintasan = 2 S∞ - a

2

412

2

2

43

1

22

12

a

r

a

= 14

Jadi panjang lintasan yang dilalui bola adalah14 m


2. An elastic ball is drop from the height of 2m. Every time it bounce from the floor, it has ¾ of the previous height. How long is the route that will be passed by the ball until it stop?


Look at the picture!The ball is drop from A, so AB is passed only once. Then CD, EF, etc is passed twice. The route is in geometric series with a = 3 and r = ¾ the length of the route is= 2 S∞ - a

2

412

2

2

43

1

22

12

a

r

a

= 14

So, the route length that pass by the ball is 14 m

Barisan dan deret 1 bilingual

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