Using Sigma Notation to represent Finite Geometric Series

📌 Example 1.
Express the geometric series, 7+14+28+56+112+224, using summation notation.
✍ Solution:
Write a formula: an=a1r(n-1)
List the information: a1=7

r=14/7
n=6 (use 6 on the Σ symbol, but leave it unknown in the argument.)
an=?

Substitute and solve: an=7⋅2(n-1).
Apply the Sigma notation: ∑6n=1 (7⋅2(n-1) )
Therefore, 7+14+28+56+112+224=∑6n=1 (7⋅2(n-1) ) .

📌 Ex2. Write each geometric series in sigma notation.
📌 Ex2a. 3+12+48+…+3072.
✍ Solution:
Find the common ratio.

12÷3=4
48÷12=4

Next, determine the upper bound.
a4=48⋅4=192
a5=192⋅4=768
a6=768⋅4=3072

Write an explicit formula for the sequence.
an=a1r(n-1)
=3⋅4(n-1)

Therefore, in sigma notation the series 3+12+48+…+3072 can be written as ∑6n=1 3⋅4(n-1).

📌 Ex2b. 9+18+36+…+1152.
✍ Solution:
Find the common ratio.

18÷9=2
36÷18=2

Next, determine the upper bound.

a4=36⋅2=72
a5=72⋅2=144
a6=144⋅2=288
a7=288⋅2=576
a8=576⋅2=1152

Write an explicit formula for the sequence.

an=a1r(n-1)
=9⋅2(n-1)

Therefore, in sigma notation the series 9+18+36+…+1152 can be written as ∑8n=1 9⋅2(n-1).

📌 Ex2c. 50+85+144.5+…+417.605.
✍ Solution:
Find the common ratio.

85÷50=1.7
144.5÷85=1.7

Next, determine the upper bound.

a4=144.5⋅1.7=245.65
a5=245.65⋅1.7=417.605

Write an explicit formula for the sequence.

an=a1r(n-1)
=50⋅(1.7)(n-1)

Therefore, in sigma notation the series 50+85+144.5+…+417.605 can be written as ∑5n=1 50⋅(1.7)(n-1).

📌 Ex2d. ⅛-¼+½-…+8.
✍ Solution:
Find the common ratio.

-¼÷⅛=-2
½÷(-¼)=-2

Next, determine the upper bound.

a4=½⋅(-2)=-1
a5=-1⋅(-2)=2
a6=2⋅(-2)=-4
a7=-4⋅(-2)=8

Write an explicit formula for the sequence.

an=a1r(n-1)
=⅛⋅(-2)(n-1)

Therefore, in sigma notation the series ⅛+(-¼)+½+…+8 can be written as ∑7n=1 ⅛⋅(-2)(n-1).

📌 Ex2e. 0.2-1+5-…-625.
✍ Solution:
Find the common ratio.

-1÷0.2=-5
5÷-1=-5

Next, determine the upper bound.

a4=5⋅(-5)=-25
a5=-25⋅(-5)=125
a6=125⋅(-5)=-625

Write an explicit formula for the sequence.

an=a1r(n-1)
=0.2⋅(-5)(n-1)

Therefore, in sigma notation the series 0.2+(-1)+5+…+(-625) can be written as ∑6n=1 0.2⋅(-5)(n-1).

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