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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