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3.1 Bewys dat $ ootnotesize{ extstyle{igg(m{ rac{ extstyle{ extstyle{igg(m{ rac{4}{3}}} + extstyle{ rac{4}{3}} + extstyle{ rac{4}{3}} + .. - NSC Mathematics - Question 3 - 2020 - Paper 1

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3.1-Bewys-dat-$-ootnotesize{-extstyle{igg(m{-rac{-extstyle{-extstyle{igg(m{-rac{4}{3}}}-+--extstyle{-rac{4}{3}}-+--extstyle{-rac{4}{3}}-+-..-NSC Mathematics-Question 3-2020-Paper 1.png

3.1 Bewys dat $ ootnotesize{ extstyle{igg(m{ rac{ extstyle{ extstyle{igg(m{ rac{4}{3}}} + extstyle{ rac{4}{3}} + extstyle{ rac{4}{3}} + ... + extstyle{ rac{4}... show full transcript

Worked Solution & Example Answer:3.1 Bewys dat $ ootnotesize{ extstyle{igg(m{ rac{ extstyle{ extstyle{igg(m{ rac{4}{3}}} + extstyle{ rac{4}{3}} + extstyle{ rac{4}{3}} + .. - NSC Mathematics - Question 3 - 2020 - Paper 1

Step 1

Bewys dat $\sum_{k=1}^{\infty} \frac{4}{3^k}$ 'n konvergerende meetkundige reeks is.

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Answer

To prove that the series converges, we will use the formula for the sum of an infinite geometric series, which is given by:

S=a1rS = \frac{a}{1 - r}

where:

  • (a) is the first term of the series,
  • (r) is the common ratio of the series.

From the series ootnotesize{\sum_{k=1}^{\infty} \frac{4}{3^k}}:

  • The first term, (a = \frac{4}{3} ), when ( k = 1).
  • The common ratio, ( r = \frac{4/3^k}{4/3^{k-1}} = \frac{1}{3} ) for ( k \geq 1 ).

Since the magnitude of the common ratio must be less than 1 for convergence:

r<1    1<13<1.|r| < 1 \implies -1 < \frac{1}{3} < 1.

Thus, the series converges.

Step 2

Indien $\sum_{k=p}^{\infty} \frac{4}{3^k} = \frac{2}{9}, bepaal die waarde van p.$

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Answer

We start with the general term for the sum of the series from ( k = p ):

S=k=parkp,S = \sum_{k=p}^{\infty} a r^{k - p}, where ( a = \frac{4}{3^p} ) and ( r = \frac{1}{3}.)

Thus, the sum becomes:

S=43p113=43p23=423p1=23p1.S = \frac{\frac{4}{3^p}}{1 - \frac{1}{3}} = \frac{\frac{4}{3^p}}{\frac{2}{3}} = \frac{4}{2 \cdot 3^{p-1}} = \frac{2}{3^{p-1}}.

Setting this equal to ( \frac{2}{9} ):

23p1=29.\frac{2}{3^{p-1}} = \frac{2}{9}.

By simplifying both sides, we find:

3p1=9=32    p1=2    p=3.3^{p-1} = 9 = 3^2 \implies p - 1 = 2 \implies p = 3.

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