Question

Drag the tiles to the correct boxes to complete the pairs. Not all tiles will be used.

Answer 1

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

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:

$\sum\limits^4_0 3(5)^n$

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:

$\sum\limits^4_0 3(5)^n = 3 + 15 +75 + 375 +1875$

$\sum\limits^4_0 4(8)^n$

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:

$\sum\limits^4_0 4(8)^n = 4 + 32 + 256+ 2048 + 16384$

$\sum\limits^4_0 2(3)^n$

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:

$\sum\limits^4_0 2(3)^n = 2 + 6 + 18 + 54 + 162$

$\sum\limits^4_0 5(3)^n$

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

$\sum\limits^4_0 5(3)^n = 5 + 15 + 45 + 135 + 405$

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

Read more about sigma notation at:

brainly.com/question/542712

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