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Match each series with the equivalent series written in sigma notation.

[tex]\[
\begin{array}{l}
\sum_{n=0}^4 2(9)^n \quad \sum_{n=0}^4 4(8)^n \quad \sum_{n=0}^4 3(5)^n \quad \sum_{n=0}^4 3(2)^n \quad \sum_{n=0}^4 5(3)^n \\
\sum_{n=0}^4 2(3)^n \quad \sum_{n=0}^4 3(4)^n \quad \sum_{n=0}^4 4(4)^n \\
3+15+75+375+1{,}875 \longleftrightarrow \\
2+6+18+54+162 \\
4+32+256+2{,}048+16{,}384 \longleftrightarrow \\
3+12+48+192+768 \longleftrightarrow \\
\end{array}
\][/tex]

- [tex]\(3 + 15 + 75 + 375 + 1{,}875 \longleftrightarrow \sum_{n=0}^4 3(5)^n\)[/tex]
- [tex]\(2 + 6 + 18 + 54 + 162 \longleftrightarrow \sum_{n=0}^4 2(3)^n\)[/tex]
- [tex]\(4 + 32 + 256 + 2{,}048 + 16{,}384 \longleftrightarrow \sum_{n=0}^4 4(8)^n\)[/tex]
- [tex]\(3 + 12 + 48 + 192 + 768 \longleftrightarrow \sum_{n=0}^4 3(4)^n\)[/tex]



Answer :

GinnyAnswer
Let's break down the process of matching each given series with its equivalent series written in sigma notation one by one:

### Series 1: [tex]\(3 + 15 + 75 + 375 + 1,875\)[/tex]
Observe that each term is the previous term multiplied by 5:
- [tex]\(3 = 3 \cdot 5^0\)[/tex]
- [tex]\(15 = 3 \cdot 5^1\)[/tex]
- [tex]\(75 = 3 \cdot 5^2\)[/tex]
- [tex]\(375 = 3 \cdot 5^3\)[/tex]
- [tex]\(1,875 = 3 \cdot 5^4\)[/tex]

This series can be represented by the sigma notation:
[tex]\[ \sum_{n=0}^4 3(5)^n \][/tex]

### Series 2: [tex]\(2 + 6 + 18 + 54 + 162\)[/tex]
Observe that each term is the previous term multiplied by 3:
- [tex]\(2 = 2 \cdot 3^0\)[/tex]
- [tex]\(6 = 2 \cdot 3^1\)[/tex]
- [tex]\(18 = 2 \cdot 3^2\)[/tex]
- [tex]\(54 = 2 \cdot 3^3\)[/tex]
- [tex]\(162 = 2 \cdot 3^4\)[/tex]

This series can be represented by the sigma notation:
[tex]\[ \sum_{n=0}^4 2(3)^n \][/tex]

### Series 3: [tex]\(4 + 32 + 256 + 2,048 + 16,384\)[/tex]
Observe that each term is the previous term multiplied by 8:
- [tex]\(4 = 4 \cdot 8^0\)[/tex]
- [tex]\(32 = 4 \cdot 8^1\)[/tex]
- [tex]\(256 = 4 \cdot 8^2\)[/tex]
- [tex]\(2,048 = 4 \cdot 8^3\)[/tex]
- [tex]\(16,384 = 4 \cdot 8^4\)[/tex]

This series can be represented by the sigma notation:
[tex]\[ \sum_{n=0}^4 4(8)^n \][/tex]

### Series 4: [tex]\(3 + 12 + 48 + 192 + 768\)[/tex]
Observe that each term is the previous term multiplied by 4:
- [tex]\(3 = 3 \cdot 4^0\)[/tex]
- [tex]\(12 = 3 \cdot 4^1\)[/tex]
- [tex]\(48 = 3 \cdot 4^2\)[/tex]
- [tex]\(192 = 3 \cdot 4^3\)[/tex]
- [tex]\(768 = 3 \cdot 4^4\)[/tex]

This series can be represented by the sigma notation:
[tex]\[ \sum_{n=0}^4 3(4)^n \][/tex]

So, the final matching of each series with its corresponding sigma notation is as follows:

1. [tex]\(3 + 15 + 75 + 375 + 1,875\)[/tex] corresponds to [tex]\(\sum_{n=0}^4 3(5)^n\)[/tex]
2. [tex]\(2 + 6 + 18 + 54 + 162\)[/tex] corresponds to [tex]\(\sum_{n=0}^4 2(3)^n\)[/tex]
3. [tex]\(4 + 32 + 256 + 2,048 + 16,384\)[/tex] corresponds to [tex]\(\sum_{n=0}^4 4(8)^n\)[/tex]
4. [tex]\(3 + 12 + 48 + 192 + 768\)[/tex] corresponds to [tex]\(\sum_{n=0}^4 3(4)^n\)[/tex]

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