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Match each series with the equivalent series written in sigma notation

Drag The Tiles To The Correct Boxes To Complete The Pairs Not All Tiles Will Be Used Match Each Series With The Equivalent Series Written In Sigma Notation class=

Sagot :

MrRoyalidn

The series and the sigma notations are[tex]\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875[/tex], [tex]\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384[/tex], [tex]\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162[/tex] and [tex]\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405[/tex]

How to match each series with the equivalent series written in sigma notation?

To do this, we simply expand each sigma notation.

So, we have:

[tex]\sum\limits^4_0 3(5)^n[/tex]

Next, we set n = 0 to 4.

So, we have:

3(5)^0 = 3

3(5)^1 = 15

3(5)^2 = 75

3(5)^3 = 375

3(5)^4 = 1875

So, we have:

[tex]\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875[/tex]

[tex]\sum\limits^4_0 4(8)^n[/tex]

Next, we set n = 0 to 4.

So, we have:

4(8)^0 = 4

4(8)^1 = 32

4(8)^2 = 256

4(8)^3 = 2048

4(8)^4 = 16384

So, we have:

[tex]\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384[/tex]

[tex]\sum\limits^4_0 2(3)^n[/tex]

Next, we set n = 0 to 4.

So, we have:

2(3)^0 = 2

2(3)^1 = 6

2(3)^2 = 18

2(3)^3 = 54

2(3)^4 = 162

So, we have:

[tex]\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162[/tex]

[tex]\sum\limits^4_0 5(3)^n[/tex]

Next, we set n = 0 to 4.

So, we have:

5(3)^0 = 5

5(3)^1 = 15

5(3)^2 = 45

5(3)^3 = 135

5(3)^4 = 405

[tex]\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405[/tex]

Hence, the series and the sigma notations are[tex]\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875[/tex], [tex]\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384[/tex], [tex]\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162[/tex] and [tex]\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405[/tex]

Read more about sigma notation at:

https://brainly.com/question/542712

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