Kumiswa, Imfundo yesibili nezikole
Inchubekelembili-Jomethri. ISIBONELO esinqumweni
Cabanga ilandelana.
7 28 112 448 1792 ...
sikhombisa Kuyabonakala ukuthi ukubaluleka yimiphi izakhi zalo ezingaphezu kuka owedlule ncamashi kane. Ngakho, lolu chungechunge kuyinto inchubekelembili.
inchubekelembili yejeyomethri ngokuthi ukulandelana esingapheliyo izinombolo, isici esiyinhloko okuyinto ukuthi inombolo elandelayo is etholakala ngenhla ngokuvele azalane ngenombolo ethile. Lokhu kuboniswa ifomula elandelayo.
a z ukubeka +1 = a z · q , lapho z - nenani element ekhethiwe.
Ngakho, z ∈ N.
Isikhathi lapho isikole sifundwa inchubekelembili weJiyomethri - grade 9. Izibonelo kuzosiza baqonde umqondo:
0.25 0.125 0.0625 ...
18 Februwari 6 ...
Kususelwa kule formula, ukuqhubeka zihilela kungatholwa kanje:
Ngiphakathi q, noma b z awukwazi ukuba uziro. Futhi, ngamunye izingxenye uchungechunge lwezinombolo yokuragela phambili kufanele iphathe ukuba uziro.
Ngakho, ukubona inombolo eduze yenombolo, uphindaphinde yokugcina ngo q.
Ukuze uchaze Kuleli qoqo, kuzomele ucacise element lokuqala ke futhi zifana. Ngemva kwalokho kungenzeka ukuthola noma yimaphi amalungu alandelayo futhi lemali yabo.
zinhlobo
Kuye q kanye 1, Kuleli qoqo ihlukaniswe izinhlobo eziningana:
- Uma 1, futhi q mkhulu kwelilodwa ke ukulandelana - okwandisa ngamunye isici ezilandelanayo umkhuba oqhubekayo wokubuka weJiyomethri. Izibonelo zazo zi.
Isibonelo: 1 = 3, q = 2 - mkhulu kunobunye, kokubili imingcele.
Khona-ke ukulandelana kweenomboro singatlolwa bunjesi:
3 6 12 24 48 ...
- Uma | q | engaphansi, ie-ke okulingana ukubuyabuyelela ukwahlukana, ukuqhubeka nezimo ezifanayo - ngincipha inchubekelembili weJiyomethri. Izibonelo zazo zi.
Isibonelo: 1 = 6, q = 1/3 - a 1 mkhulu kweyodwa, q - kancane.
Khona-ke ukulandelana kweenomboro kungenziwa ebhaliwe ngale ndlela:
Juni 2 2/3 ... - izakhi yimuphi isici ngaphezulu uyalilandela, kuyinto izikhathi 3.
- Kushintshana. Uma q <0, izibonakaliso Izinamba kushintshana ukulandelana njalo kungakhathaliseki a 1, futhi izingxenye iyiphi sokukhuphula noma ukwehlisa.
Isibonelo: 1 = -3, q = -2 - kukhona kokubili esingaphansi zero.
Khona-ke ukulandelana kweenomboro singatlolwa bunjesi:
3, 6, -12, 24, ...
ifomula
Ukuze ukusetshenziswa elula, kukhona Progressions abaningi be-Jomethri amafomula:
- Formula z-th eside. It ivumela ukubala kwalesi sakhi inombolo letsite ngaphandle kokubala izinombolo odlule.
Isibonelo: q = 3, a = 1 4. edingekayo ukubala yesine isici inchubekelembili.
Isixazululo: a = 4 4 3 · 4-1 · 3 = 4 3 = 4 · 27 = 108.
- Isamba imisuka, ogama inombolo ilingana z. It ivumela sibalo isamba zonke izakhi ukulandelana a z kukonke.
≠ 0, ngaleyo ndlela, q akuyona 1 - (q 1) Njengoba (1- q) uma zihilela ke.
Qaphela: uma q = 1, ke ukuqhubeka ngabe emelelwa eziningi bangaphezi ukuphinda inombolo.
Inani exponentially izibonelo: a 1 = 2, q = -2. Bala S 5.
Isixazululo: S 5 = 22 - ukubala ifomula.
- Inani uma | q | <1 nalapho z ivame infinity.
Isibonelo: 1 = 2, q = 0.5. Thola isamba.
Isixazululo: S z = 2 x = 4
Uma sibona isamba amalungu amaningana bhukwana, uzobona ukuthi ngempela izinikele ezine.
S z = 1 + 2 + 0.5 + 0.25 + 0,125 + 0.0625 = 3.9375 4
Ezinye izakhiwo:
- Sici impahla. Uma lokhu Ngoba kunesithembiso nganoma yisiphi z ke unikezwa uchungechunge zezinombolo - umkhuba oqhubekayo wokubuka Jomethri:
a z 2 = A z -1 · A z +1
- Libuye isikwele kwanoma iyiphi inombolo exponentially esebenzisa kwalokho segcekeni we ezinye izinombolo ezimbili yimuphi irowu unikezwa, uma equidistant kusukela isici.
2 a z = a z - t 2 + a z + t 2 lapho t - ibanga phakathi lezi zinombolo.
- Izakhi kuhluke izikhathi q.
- I logarithms izici phambili yakha kanye umkhuba oqhubekayo wokubuka, kodwa izibalo, okungukuthi, ngamunye wabo kakhulu kunaleso langaphambilini inombolo ethile.
Izibonelo ezinye izinkinga classical
Ukuze uqonde kangcono ukuthi yini ukuqhubeka weJiyomethri, ne izibonelo isinqumo grade 9 kungasiza.
- Imigomo nemibandela: 1 = 3, 3 = 48. Thola q.
Isixazululo: ngamunye isici ilandelana aminingi ukudlula q odlule isikhathi. Kuyadingeka ukuba baveze ezinye izakhi nakwezinye nge zifana.
Ngenxa yalokho, i-3 = q 2 u 1
Lapho abambele q = 4
- Izimo: 2 = 6, a = 3 12. Bala S 6.
Isixazululo: Ukuze wenze lokhu, lwanele ukuthola q, element lokuqala futhi esikhundleni ku ifomula.
3 = q · 2 consequently q = 2
2 = q · A 1, ngakho a = 1 3
S = 6 189
- · A 1 = 10, q = -2. Thola isici sesine phambili.
Isixazululo: kwanele ukuveza isici sesine ngokusebenzisa kuqala futhi ngokusebenzisa zifana.
4 3 = q u = 1 -80
Ngokwesibonelo Isicelo:
- Bank iklayenti uye waba negalelo isamba esingama-ruble angu 10,000, unyaka ngamunye iklayenti lemali oyinhloko niyozenezelelwa 6% ngaphansi okuyinto Nokho. Yimalini ku-akhawunti emva kweminyaka 4?
Isixazululo: Inani kokuqala elilingana ama-ruble ayizigidi 10. Ngakho, ngemva konyaka kwemali kwi-akhawunti kuyoba inani elilingana 10000 + 10000 = 10000 · 0,06 · 1.06
Ngakho, inani ku-akhawunti ngisho ngemva konyaka owodwa siyofezwa kanje:
(10000 · 1.06) · 10000 · 0,06 + 1,06 = 1,06 · 1.06 · 10000
Okungukuthi, unyaka ngamunye inani senyukela izikhathi 1.06. Ngakho, ukuze uthole inombolo ye-akhawunti emva kweminyaka 4, lwanele ukuthola wesine isici inchubekelembili onikelwa isici sokuqala elilingana ayizinkulungwane 10, kanye zihilela elilingana 1.06.
S = 1.06 · 1.06 · 1.06 · 1.06 · 10000 = 12625
Izibonelo zezinkinga e kathisha yesamba:
Ngo izinkinga ezihlukahlukene usebenzisa inchubekelembili weJiyomethri. Isibonelo ekutholeni isamba angathunyelwa kanje:
a 1 = 4, q = 2, ukubala S 5.
Isixazululo: yonke idatha ezidingekile sokubalwa zaziwa, umane bamelele wabayisa ifomula.
S 5 = 124
- 2 = 6, a = 3 18. Bala isamba izakhi eziyisithupha zokuqala.
isixazululo:
I Geom. ngentuthuko elementi elandelayo kunangendlela izikhathi ezidlule q ngamunye, okungukuthi, ukubala inani odinga ukukwazi element a 1 kanye q zifana.
2 · q = 3
q = 3
Ngokufanayo, isidingo ukuthola 1, 2 futhi engazi q.
a 1 · q = 2
a 1 = 2
Bese lwanele shintsha idatha eyaziwa ku lemali ifomula.
S 6 = 728.
Similar articles
Trending Now