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Inleiding

In graad 10 het jy geleer van rekenkundige rye, waar die verskil tussen opeenvolgende terme konstant was. In hierdie hoofstuk leer ons van kwadratiese rye.

Wat is 'n kwadratiese ry ?

Kwadratiese ry

'n Kwadratiese ry is 'n ry waar die tweede verskille tussen opeenvolgende terme met dieselfde hoeveelheid verskil. Dit word 'n gemene tweede verskil genoem.

Byvoorbeeld

1 ; 2 ; 4 ; 7 ; 11 ; ...

is 'n kwadratiese ry. Kom ons stel vas hoekom ...

Indien ons die verskil tussen opeenvolgende terme neem, is

a 2 - a 1 = 2 - 1 = 1 a 3 - a 2 = 4 - 2 = 2 a 4 - a 3 = 7 - 4 = 3 a 5 - a 4 = 11 - 7 = 4

dan werk ons die tweede verskille uit, wat bloot gekry word deur die verskille tussen opeenvolgende verskille { 1 ; 2 ; 3 ; 4 ; ... } te neem:

2 - 1 = 1 3 - 2 = 1 4 - 3 = 1 ...

Ons sien dan dat die tweede verskille gelyk is aan "1". Dus is [link] 'n kwadratiese ry .

Let op dat die verskille tussen opeenvolgende terme (met ander woorde, die eerste verskille) van 'n kwadratiese ry, 'n ry vorm waar daar 'n konstante verskil is tussen opeenvolgende terme. In die voorbeeld hier bo, het die ry { 1 ; 2 ; 3 ; 4 ; ... }, wat gevorm is die die verskille tussen opeenvolgende terme van [link] te neem, 'n linêere formule van die vorm a x + b .

Kwadratiese rye

Die volgende is ook voorbeelde van kwadratiese rye:

3 ; 6 ; 10 ; 15 ; 21 ; ... 4 ; 9 ; 16 ; 25 ; 36 ; ... 7 ; 17 ; 31 ; 49 ; 71 ; ... 2 ; 10 ; 26 ; 50 ; 82 ; ... 31 ; 30 ; 27 ; 22 ; 15 ; ...

Kan jy die gemene tweede verskille vir elk van die voorbeelde hier bo bereken?

Skryf neer die volgende twee terme en vind 'n formule vir die n de term in die ry 5 , 12 , 23 , 38 , . . . , . . . ,

  1. i.e. 7 , 11 , 15

  2. die tweede verskil is 4.

    As ons die ry voortsit, sal die verskille tussen terme die volgende wees:

    15 + 4 = 19

    19 + 4 = 23

  3. Dus sal die volgende twee terme in die reeks die volgende wees:

    38 + 19 = 57

    57 + 23 = 80

    Dus sal die ry die volgende wees: 5 , 12 , 23 , 38 , 57 , 80

  4. Ons weet dat die tweede verskil 4 is. Die begin van die formule sal dus 2 n 2 wees.

  5. Indien n = 1 , moet jy die volgende waarde in die ry kry, wat "5" vir hierdie spesifieke ry is. Die verskil tussen 2 n 2 = 2 en die oorspronklike getal (5) is 3, wat lei tot n + 2 .

    Kyk of dit werk vir die tweede terme, d.i. wanneer n = 2 .

    Dan is 2 n 2 = 8 . Die verskil tussen term twee en (12) en 8 is 4, wat geskryf kan word as n + 2 .

    Dus vir die ry 5 , 12 , 23 , 38 , . . . is die formule vir die n de term 2 n 2 + n + 2 .

Algemene geval

Indien die ry kwadraties is, moet die n de term T n = a n 2 + b n + c wees

TERME a + b + c 4 a + 2 b + c 9 a + 3 b + c
1 ste verskil 3 a + b 5 a + b 7 a + b
2 de verskil 2 a 2 a

In elke geval is die tweede verskil 2 a . Hierdie feit kan gebruik word om a te vind, dan b en dan c .

Die volgende ry is kwadraties: 8 , 22 , 42 , 68 , . . . Vind die formule.

  1. TERME 8 22 42 68
    1 ste verskil 14 20 26
    2 de verskil 6 6 6
  2. Dan is 2 a = 6 wat gee a = 3 En 3 a + b = 14 9 + b = 14 b = 5 En a + b + c = 8 3 + 5 + c = 8 c = 0
  3. Die formule is dus:     n de t e r m = 3 n 2 + 5 n

  4. Vir

    n = 1 , T 1 = 3 ( 1 ) 2 + 5 ( 1 ) = 8 n = 2 , T 2 = 3 ( 2 ) 2 + 5 ( 2 ) = 22 n = 3 , T 3 = 3 ( 3 ) 2 + 5 ( 3 ) = 42

Bepaling van die n de -term van 'n kwadratiese ry

Laat die n d e -term vir 'n kwadratiese ry gegee word deur

a n = A · n 2 + B · n + C

waar A , B and C konstantes is wat bepaal moet word.

a n = A · n 2 + B · n + C a 1 = A ( 1 ) 2 + B ( 1 ) + C = A + B + C a 2 = A ( 2 ) 2 + B ( 2 ) + C = 4 A + 2 B + C a 3 = A ( 3 ) 2 + B ( 3 ) + C = 9 A + 3 B + C
Laat d = a 2 - a 1 d = 3 A + B
B = d - 3 A

Die gemene tweede verskil word gekry vanaf

D = ( a 3 - a 2 ) - ( a 2 - a 1 ) = ( 5 A + B ) - ( 3 A + B ) = 2 A
A = D 2

Dus, vanuit [link] ,

B = d - 3 2 · D

Vanuit [link] ,

C = a 1 - ( A + B ) = a 1 - D 2 - d + 3 2 · D
C = a 1 + D - d

Uiteindelik word die algemene formule vir die n d e term van 'n kwadratiese ry gegee deur

a n = D 2 · n 2 + ( d - 3 2 D ) · n + ( a 1 - d + D )

Bestudeer die volgende patroon: 1; 7; 19; 37; 61; ...

  1. Wat is die volgende getal in die ry?
  2. Gebruik veranderlikes om 'n algebraïese formula op te stel wat die patroon veralgemeen.
  3. Wat sal die 100 ste term van die ry wees?
  1. Die getalle vermeerder met veelvoude van 6

    1 + 6 ( 1 ) = 7 , dan is 7 + 6 ( 2 ) = 19

    19 + 6 ( 3 ) = 37 , dan is 37 + 6 ( 4 ) = 61

    Dus is 61 + 6 ( 5 ) = 91

    Die volgende getal in die ry is 91.

  2. TERME 1 7 19 37 61
    1 ste verskil 6 12 18 24
    2 de verskil 6 6 6 6

    Die patroon sal 'n kwadratiese patroon opbring, aangesien die tweede verskille konstant is.

    Dus is a n 2 + b n + c = y

    Vir die eerste term: n = 1 , dan is y = 1

    Vir die tweede term: n = 2 , dan is y = 7

    Vir die derde term: n = 3 , dan is y = 19

    ensovoorts....

  3. a + b + c = 1 4 a + 2 b + c = 7 9 a + 3 b + c = 19
  4. verg. ( 2 ) - verg. ( 1 ) : 3 a + b = 6 verg. ( 3 ) - verg. ( 2 ) : 5 a + b = 12 verg. ( 5 ) - verg. ( 4 ) : 2 a = 6 a = 3 , b = - 3 e n c = 1
  5. Die algemene formule vir die patroon is 3 n 2 - 3 n + 1

  6. Vervang n met 100:

    3 ( 100 ) 2 - 3 ( 100 ) + 1 = 29 701

    Die waarde van die 100 ste term is 29 701.

Teken 'n grafiek van die terme van 'n kwadratiese ry

Die plot van a n vs. n lewer 'n paraboliese grafiek vir 'n kwadratiese ry,

gegee die kwadratiese ry

3 ; 6 ; 10 ; 15 ; 21 ; ...

Indien ons elke van die terme teenoor die ooreenstemmende indeks teken, kry ons die grafiek van 'n parabool.

Oefeninge

  1. Vind die eerste 5 terme van die kwadratiese ry gedefinieer deur:
    a n = n 2 + 2 n + 1
  2. Bepaal watter van die volgende rye kwadraties is deur die gemene tweede verskille te bereken:
    1. 6 ; 9 ; 14 ; 21 ; 30 ; ...
    2. 1 ; 7 ; 17 ; 31 ; 49 ; ...
    3. 8 ; 17 ; 32 ; 53 ; 80 ; ...
    4. 9 ; 26 ; 51 ; 84 ; 125 ; ...
    5. 2 ; 20 ; 50 ; 92 ; 146 ; ...
    6. 5 ; 19 ; 41 ; 71 ; 109 ; ...
    7. 2 ; 6 ; 10 ; 14 ; 18 ; ...
    8. 3 ; 9 ; 15 ; 21 ; 27 ; ...
    9. 10 ; 24 ; 44 ; 70 ; 102 ; ...
    10. 1 ; 2 , 5 ; 5 ; 8 , 5 ; 13 ; ...
    11. 2 , 5 ; 6 ; 10 , 5 ; 16 ; 22 , 5 ; ...
    12. 0 , 5 ; 9 ; 20 , 5 ; 35 ; 52 , 5 ; ...
  3. Gegee a n = 2 n 2 , vind die waarde van n , a n = 242
  4. Gegee a n = ( n - 4 ) 2 , vind vir watter waarde van n , a n = 36
  5. Gegee a n = n 2 + 4 , vind die waarde van n , a n = 85
  6. Gegee a n = 3 n 2 , vind a 11
  7. Gegee a n = 7 n 2 + 4 n , vind a 9
  8. Gegee a n = 4 n 2 + 3 n - 1 , vind a 5
  9. Gegee a n = 1 , 5 n 2 , vind a 10
  10. Vir elke van die kwadratiese rye, vind die gemene tweede verskil, die formule vir die algemene term en gebruik dan die formule om a 100 te vind.
    1. 4 , 7 , 12 , 19 , 28 , ...
    2. 2 , 8 , 18 , 32 , 50 , ...
    3. 7 , 13 , 23 , 37 , 55 , ...
    4. 5 , 14 , 29 , 50 , 77 , ...
    5. 7 , 22 , 47 , 82 , 127 , ...
    6. 3 , 10 , 21 , 36 , 55 , ...
    7. 3 , 7 , 13 , 21 , 31 , ...
    8. 3 , 9 , 17 , 27 , 39 , ...

Questions & Answers

can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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a perfect square v²+2v+_
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algebra 2 Inequalities:If equation 2 = 0 it is an open set?
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or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
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Kristine 2*2*2=8
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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J, combine like terms 7x-4y
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China
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
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Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
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silver nanoparticles could handle the job?
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Siyavula textbooks: wiskunde (graad 11). OpenStax CNX. Sep 20, 2011 Download for free at http://cnx.org/content/col11339/1.4
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