Eisimpleirean de Cheistean a’ Deasbad Sreathan Geoimeatrach
Tha sreathan geoimeatrach nam bun-bheachd deatamach ann am matamataig, a’ nochdadh gu tric ann an diofar sheòrsaichean dhuilgheadasan, nam measg deuchainnean sgoile, deuchainnean inntrigidh colaiste, agus eadhon deuchainnean àbhaisteach mar an SAT no GRE. Bidh tuigse mhionaideach air sreathan geoimeatrach gar cuideachadh le bhith a’ fuasgladh dhuilgheadasan gu h-èifeachdach. Còmhdaichidh an artaigil seo grunn eisimpleirean de dhuilgheadasan agus bruidhnidh e air sreathan geoimeatrach gu mionaideach.
A’ Tuigsinn Sreathan Geoimeatrach
Is e sreath geoimeatrach sreath anns a bheil gach teirm air fhaighinn le bhith ag iomadachadh an teirm roimhe le àireamh stèidhichte ris an canar an co-mheas (co-mheas cumanta, mar as trice air a chomharrachadh leis an litir \(r\)). San fharsaingeachd, faodar sreath geoimeatrach a sgrìobhadh mar:
\[
a, ar, ar^2, ar^3, \ldots
\]
Càite:
– ’S e \(a\) a’ chiad teirm
– Is e \(r\) co-mheas an t-sreath
Ma tha \( |r| < 1 \), tha feart inntinneach aig sreathan geoimeatrach neo-chrìochnach de cho-chruinneachadh. Tha mòran thagraidhean practaigeach aig sreathan geoimeatrach ann an diofar raointean leithid fiosaig, eaconamas agus bith-eòlas.
Foirmle Sreath Geoimeatrach An nmh teirm de shreath geoimeatrach Faodar an nmh teirm de shreath geoimeatrach obrachadh a-mach leis an fhoirmle: \[ U_n = a \cdot r^{n-1} \] Suim a’ Chiad n Teirmean de Shreath Geoimeatrach Faodar suim a’ chiad \(n\) teirmean de shreath geoimeatrach (Sn) obrachadh a-mach leis an fhoirmle: \[ S_n = a \frac{1 - r^n}{1 - r}, \quad \text{airson } r \neq 1 \] \[ S_n = na, \quad \text{airson } r = 1 \] Suim Neo-chrìochnach de Shreath Geoimeatrach Ma tha \(|r| < 1\), tha an t-suim aig sreath geoimeatrach neo-chrìochnach: \[ S_{\infty} = \frac{a}{1 - r} \] Eisimpleir de Cheistean agus Deasbadan Seo eisimpleirean de cheistean sreathan geoimeatrach còmhla ris na deasbadan aca: Eisimpleir de Cheist 1: A’ tomhas an nmh Teirm Ceist: Sreath geoimeatrach leis a’ chiad teirm \(a = 5\) agus an co-mheas cumanta \(r = 3\). Obraich a-mach an 6mh teirm den t-sreath. Fuasgladh: A’ cleachdadh foirmle an nmh teirm: [U_6 = a ⋅r^{(6-1)} = 5 ⋅3^5 = 5 ⋅243 = 1215 \] Mar sin, is e 1215 an 6mh teirm den t-sreath. Eisimpleir Ceist 2: A’ tomhas suim a’ chiad n teirmean Ceist: Obraich a-mach suim a’ chiad 4 teirmean de shreath geoimeatrach leis a’ chiad teirm _(a = 2)_ agus an co-mheas _(r = \frac{1}{2}__). Deasbad: A’ cleachdadh na foirmle airson suim a’ chiad theirmean _(n_): _[ S_4 = a _(1 - r^4}{1 - r} = 2 _(1 - (1}{2)^4}{1 - \frac{1}{2}} = 2 _(1 - \frac{1}{16}}{\frac{1}{2}} = 2 _(15}{16}}{\frac{1}{2}} = 2 _(15}{8} = 2 _(15}{8} = 3.75)_ Mar sin, is e 3.75 suim a’ chiad 4 teirmean den t-sreath. Eisimpleir 3: Suim Sreath Geoimeatrach Neo-chrìochnach Ceist: Obraich a-mach suim sreath neo-chrìochnach far a bheil \(a = 7\) agus \(r = \frac{1}{3}\). Fuasgladh: A’ cleachdadh na foirmle airson suim sreath neo-chrìochnach: \[ S_{\infty} = \frac{a}{1 - r} = \frac{7}{1 - \frac{1}{3}} = \frac{7}{\frac{2}{3}} = 7 \cdot \frac{3}{2} = \frac{21}{2} = 10.5 \] Mar sin, is e 10.5 suim an t-sreath neo-chrìochnach. Eisimpleir 4: A’ dearbhadh nan teirmean agus a’ cho-mheas de shreath Ceist: Is e 21 suim a’ chiad 3 teirmean de shreath geoimeatrach, agus is e 18 suim an 2na agus an 3mh teirm. Obraich a-mach a’ chiad teirm agus a cho-mheas. Deasbad: Ma tha a’ chiad teirm na \(a\) agus is e \(r\) an co-mheas. Bho fhiosrachadh na trioblaid, is urrainn dhuinn an dà cho-aontar a leanas a sgrìobhadh: [ a + ar + ar^2 = 21 \quad \text{(1)} \] \[ ar + ar^2 = 18 \quad \text{(2)} \] Bho cho-aontar (2), is urrainn dhuinn \(a\) a chur an cèill a thaobh \(r\): \[ a(r + r^2) = 18 \implies a = \frac{18}{r(1 + r)} \] An uairsin, cuir \(a\) a-steach do cho-aontar (1): \[ \frac{18(1)}{r(1 + r)} + \frac{18r}{r(1 + r)} + \frac{18r^2}{r(1 + r)} = 21 \] \[ \frac{18}{1 + r} + \frac{18r}{1 + r} + \frac{18r^2}{1 + r} = 21 \] \[ \frac{18 (1 + r + r^2)}{1 + r} = 21 \] \[ \frac{18 \cdot 3}{1 + r} = 21 \] \[ \frac{54}{1 + r} = 21 \] \[ 54 = 21(1 + r) \] \[ 54 = 21 + 21r \] \[ 33 = 21r \] \[ r = \frac{33}{21} = \frac{11}{7} \] Leis an luach de \(r\) aithnichte, cuir air ais e ann an luach \(a\): \[ a = \frac{18}{r(1 + r)} = \frac{18}{\frac{11}{7} (1 + \frac{11}{7})} = \frac{18}{\frac{11}{7} \cdot \frac{18}{7}} = \frac{18 \cdot 7}{11 18 = 7/11 Mar sin, is e 7/11 a’ chiad teirm (a) agus is e 11/7 an co-mheas cumanta. Co-dhùnadh Tha sreathan geoimeatrach mar aon de na bun-bheachdan matamataigeach a thathas a’ cleachdadh gu farsaing ann an diofar thagraidhean. Tha tuigse air na foirmlean bunaiteach leithid an nmh teirm, suim a’ chiad n teirmean, agus suim sreath geoimeatrach neo-chrìochnach glè chudromach gus diofar dhuilgheadasan matamataigeach co-cheangailte fhuasgladh. Le bhith a’ cleachdadh diofar eisimpleirean mar a chaidh a dheasbad san artaigil seo, is urrainn dhuinn ar comas air sreathan geoimeatrach a thuigsinn agus a chleachdadh nas fheàrr a gheurachadh.