NSC Mathematics Number patterns, sequences and series: 3.1 Prove that $\, \sum

3.1 Prove that $\, \sum_{k=1}^{\infty} 4 \cdot 3^{2-k} \,$ is a convergent geometric series - NSC Mathematics - Question 3 - 2020 - Paper 1

Question 3

$3.1-Prove-that-$\,-\sum_{k=1}^{\infty}-4-\cdot-3^{2-k}-\,$-is-a-convergent-geometric-series-NSC Mathematics-Question 3-2020-Paper 1.png$

3.1 Prove that $\, \sum_{k=1}^{\infty} 4 \cdot 3^{2-k} \,$ is a convergent geometric series. Show ALL your calculations. 3.2 If $\, \sum_{k=p}^{\infty} 4 \cdot 3^{2... show full transcript

Worked Solution & Example Answer:3.1 Prove that $\, \sum_{k=1}^{\infty} 4 \cdot 3^{2-k} \,$ is a convergent geometric series - NSC Mathematics - Question 3 - 2020 - Paper 1

Step 1

Prove that $\sum_{k=1}^{\infty} 4 \cdot 3^{2-k}$ is a convergent geometric series.

96%

114 rated

Answer

To show that the series is a convergent geometric series, we first identify its form. The general term of the series can be expressed as: $a_k = 4 \cdot 3^{2-k}.$

Next, we rewrite the term: $a_k = 4 \cdot \frac{3^2}{3^k} = 12 \cdot \left(\frac{1}{3}\right)^{k-2}.$

This shows that our first term $a = 12$ and the common ratio $r = \frac{1}{3}$ . For a geometric series to converge, the condition is that the absolute value of the common ratio must be less than 1: $|r| < 1\text{, where } r = \frac{1}{3}.$

Since $|\frac{1}{3}| < 1$ , the series is indeed convergent.

Step 2

If $\sum_{k=p}^{\infty} 4 \cdot 3^{2-k} = \frac{2}{9},$ determine the value of $p.$

99%

104 rated

Answer

The sum of an infinite geometric series is given by: $S_\infty = \frac{a}{1 - r},$ where $a$ is the first term and $r$ is the common ratio. For our series:

First term $a = 4 \cdot 3^{2-p}$ (this accounts for starting at $k = p$ ).
Common ratio $r = \frac{1}{3}$ . Thus, the sum is: $S_\infty = \frac{4 \cdot 3^{2-p}}{1 - \frac{1}{3}} = \frac{4 \cdot 3^{2-p}}{\frac{2}{3}} = 6 \cdot 3^{2-p}.$

Now, we set this equal to $\frac{2}{9}$ : $6 \cdot 3^{2-p} = \frac{2}{9}.$ Dividing both sides by 6, we have: $3^{2-p} = \frac{1}{27} = 3^{-3}.$

Equating the exponents gives: $2 - p = -3 \Rightarrow p = 5.$

The value of $p$ is therefore $5.$

3.1 Prove that $\, \sum_{k=1}^{\infty} 4 \cdot 3^{2-k} \,$ is a convergent geometric series - NSC Mathematics - Question 3 - 2020 - Paper 1

Worked Solution & Example Answer:3.1 Prove that $\, \sum_{k=1}^{\infty} 4 \cdot 3^{2-k} \,$ is a convergent geometric series - NSC Mathematics - Question 3 - 2020 - Paper 1

Prove that $\sum_{k=1}^{\infty} 4 \cdot 3^{2-k}$ is a convergent geometric series.

If $\sum_{k=p}^{\infty} 4 \cdot 3^{2-k} = \frac{2}{9},$ determine the value of $p.$

Join the NSC students using SimpleStudy...