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Cyfres Geometrig Geometric Series @mathemateg /adolygumathemateg.

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Presentation on theme: "Cyfres Geometrig Geometric Series @mathemateg /adolygumathemateg."β€” Presentation transcript:

1 Cyfres Geometrig Geometric Series @mathemateg /adolygumathemateg

2 Cyfres Geometrig Geometric Series
Mae cyfres o rifau yn gyfres geometrig os ydym yn lluosi efo’r un cysonyn i gael y rhif nesaf yn y gyfres. A sequence of numbers is a geometric sequence if we multiply by the same constant to obtain the next number in the sequence. Er enghraifft, mae’r gyfres 3, 6, 12, 24, 48, ... yn gyfres geometrig ble mae’r term cyntaf yn 3 ac mae’r gymhareb cyffredin yn 2. For example, the sequence 3, 6, 12, 24, 48, ... is a geometric sequence where the first term is 3 and the common ratio is 2. Terminoleg / Terminology π‘Ž Term cyntaf y gyfres First term of the sequence π‘Ÿ Y gymhareb cyffredin The common ratio 𝑙 Term olaf y gyfres, os oes un yn bodoli Last term of the sequence, if one exists 𝑑 𝑛 𝑛fed term y gyfres The 𝑛th term of the sequence

3 Cyfres Geometrig Geometric Series
Ar gyfer cyfres geometrig efo term cyntaf π‘Ž a chymhareb cyffredin π‘Ÿ: For a geometric sequence with first term π‘Ž and common ratio π‘Ÿ: Term cyntaf / First term 𝑑 1 =π‘Ž Ail derm / Second term 𝑑 2 =π‘Žπ‘Ÿ Trydydd term / Third term 𝑑 3 =π‘Ž π‘Ÿ 2 Pedwerydd term / Fourth term 𝑑 4 =π‘Ž π‘Ÿ 3 Nfed term / Nth term 𝑑 𝑛 =π‘Ž π‘Ÿ π‘›βˆ’1 Ar gyfer pob cyfanrif 𝑛 / For all integers 𝑛, π‘Ÿ= 𝑑 𝑛+1 𝑑 𝑛

4 Cyfres Geometrig Geometric Series
Mae’n bosib ystyried cyfanswm 𝑆 𝑛 yr 𝑛 term cyntaf mewn cyfres geometrig. The geometric series 𝑆 𝑛 is the sum of the first 𝑛 terms of a geometric sequence. Ar gyfer y gyfres geometrig 3, 6, 12, 24, 48, .... / For the geometric sequence 3, 6, 12, 24, 48, .... 𝑆 3 =3+6+12=21 𝑆 6 = =189

5 Swm 𝑛 term cyntaf cyfres geometrig
Profwch mai swm 𝑛 term cyntaf cyfres geometrig efo term cyntaf π‘Ž a chymhareb cyffredin π‘Ÿ yw 𝑆 𝑛 = π‘Ž 1βˆ’ π‘Ÿ 𝑛 1βˆ’π‘Ÿ . Prawf Term 1af: 𝑑 1 =π‘Ž 2il Derm: 𝑑 2 =π‘Žπ‘Ÿ Nfed Term: 𝑑 𝑛 =π‘Ž π‘Ÿ π‘›βˆ’1 Swm yr 𝑛 term cyntaf: 𝑆 𝑛 = 𝑑 1 + 𝑑 2 +β‹―+ 𝑑 π‘›βˆ’1 + 𝑑 𝑛 𝑆 𝑛 =π‘Ž+π‘Žπ‘Ÿ+β‹―+π‘Ž π‘Ÿ π‘›βˆ’2 +π‘Ž π‘Ÿ π‘›βˆ’1 ––––– Lluosi efo π‘Ÿ: π‘Ÿ 𝑆 𝑛 =π‘Žπ‘Ÿ+π‘Ž π‘Ÿ 2 +β‹―+π‘Ž π‘Ÿ π‘›βˆ’1 +π‘Ž π‘Ÿ 𝑛 ––––– Yn gwneud  – ο‚‚ : 𝑆 𝑛 βˆ’π‘Ÿ 𝑆 𝑛 =π‘Žβˆ’π‘Ž π‘Ÿ 𝑛 𝑆 𝑛 1βˆ’π‘Ÿ =π‘Ž 1βˆ’ π‘Ÿ 𝑛 𝑆 𝑛 = π‘Ž 1βˆ’ π‘Ÿ 𝑛 1βˆ’π‘Ÿ QED

6 The sum of the first 𝑛 terms of a geometric sequence
Prove that the sum of the first 𝑛 terms of a geometric sequence with first term π‘Ž and common ratio π‘Ÿ is given by 𝑆 𝑛 = π‘Ž 1βˆ’ π‘Ÿ 𝑛 1βˆ’π‘Ÿ . Proof 1st Term: 𝑑 1 =π‘Ž 2nd Term: 𝑑 2 =π‘Žπ‘Ÿ Nth Term: 𝑑 𝑛 =π‘Ž π‘Ÿ π‘›βˆ’1 Sum of the first 𝑛 terms: 𝑆 𝑛 = 𝑑 1 + 𝑑 2 +β‹―+ 𝑑 π‘›βˆ’1 + 𝑑 𝑛 𝑆 𝑛 =π‘Ž+π‘Žπ‘Ÿ+β‹―+π‘Ž π‘Ÿ π‘›βˆ’2 +π‘Ž π‘Ÿ π‘›βˆ’1 ––––– Multiplying by π‘Ÿ: π‘Ÿ 𝑆 𝑛 =π‘Žπ‘Ÿ+π‘Ž π‘Ÿ 2 +β‹―+π‘Ž π‘Ÿ π‘›βˆ’1 +π‘Ž π‘Ÿ 𝑛 ––––– Subtracting  – ο‚‚ : 𝑆 𝑛 βˆ’π‘Ÿ 𝑆 𝑛 =π‘Žβˆ’π‘Ž π‘Ÿ 𝑛 𝑆 𝑛 1βˆ’π‘Ÿ =π‘Ž 1βˆ’ π‘Ÿ 𝑛 𝑆 𝑛 = π‘Ž 1βˆ’ π‘Ÿ 𝑛 1βˆ’π‘Ÿ QED

7 Swm i anfeidredd Sum to infinity
Os yw π‘Ÿ <1 mae’n bosib ystyried cyfanswm 𝑆 ∞ holl dermau cyfres geometrig. If π‘Ÿ <1 we can consider the sum 𝑆 ∞ of all terms of a geometric sequence. O’r prawf gynt, gwyddom fod 𝑆 𝑛 = π‘Ž 1βˆ’ π‘Ÿ 𝑛 1βˆ’π‘Ÿ . From the previous proof, we know that 𝑆 𝑛 = π‘Ž 1βˆ’ π‘Ÿ 𝑛 1βˆ’π‘Ÿ . Os yw π‘Ÿ <1, hynny yw βˆ’1<π‘Ÿ<1, yna bydd π‘Ÿ 𝑛 yn lleihau wrth i 𝑛 gynyddu. If π‘Ÿ <1, that is βˆ’1<π‘Ÿ<1, then as 𝑛 increases π‘Ÿ 𝑛 decreases. Dywedwn bod π‘Ÿ 𝑛 β†’0 fel mae π‘›β†’βˆž. We say that π‘Ÿ 𝑛 β†’0 as π‘›β†’βˆž. Felly os yw π‘Ÿ <1 mae 𝑆 ∞ = π‘Ž 1βˆ’π‘Ÿ . Therefore if π‘Ÿ <1 we have 𝑆 ∞ = π‘Ž 1βˆ’π‘Ÿ .


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