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Besku die meetkundige reeks: 4 + 2 + 1 + \frac{1}{2} + .. - NSC Mathematics - Question 2 - 2024 - Paper 1

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Besku die meetkundige reeks: 4 + 2 + 1 + \frac{1}{2} + ... 2.1.1 Konvergeer die reeks? Motiveer jou antwoord. 2.1.2 Bereken $S_n$. Gegee: $\sum_{r=1}^{10} 3^{r} =... show full transcript

Worked Solution & Example Answer:Besku die meetkundige reeks: 4 + 2 + 1 + \frac{1}{2} + .. - NSC Mathematics - Question 2 - 2024 - Paper 1

Step 1

Konvergeer die reeks? Motiveer jou antwoord.

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Answer

Die reeks konvergeer indien die gemene verhouding rr in die meetkundige reeks tussen -1 en 1 val. Hier is die eerste term a=4a = 4 en die gemene verhouding r=24=12r = \frac{2}{4} = \frac{1}{2}.

Hierdie waarde van rr is binne die interval (1,1)(-1, 1), dus convergeer die reeks.

Step 2

Bereken $S_n$.

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Die formule vir die som van 'n meetkundige reeks is gegee deur: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r} Hier is a=4a = 4 en r=12r = \frac{1}{2}.

Substitusie in die formule: Sn=4(1(12)n)112=4(1(12)n)12=8(1(12)n).S_n = \frac{4(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{4(1 - (\frac{1}{2})^n)}{\frac{1}{2}} = 8(1 - (\frac{1}{2})^n).

Step 3

Bereken die waarde van $k$.

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Die som van die reeks is gegee as: \sum_{r=1}^{10} 3^{r} = 29 520.\

Hierdie som kan weergegee word as: Sn=a(1rn)/(1r),S_n = a(1 - r^{n}) / (1 - r), waar a=31=3a = 3^{1} = 3 en r=3.r = 3. Dit impliseer dat: 29520=3(1310k)/(13).29 520 = 3(1 - 3^{10 - k}) / (1 - 3).

Die kontantsom van die som is 311k=29520imes2=59040.3^{11 - k} = 29 520 imes 2 = 59 040.

Sien ons dat 310=590493^{10} = 59 049.

Dus, k=11n=1110=1.k = 11 - n = 11 - 10 = 1.

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