**Evaluating some Sums which are written in Sigma Notation**

**Sum of a Finite Geometric Series**

Geometric series and fractal geometry are connected and have many applications in the realms of physics, engineering, and economics.

A geometric series is a geometric sequence whose terms are added.

The general form for a geometric series can be expressed using summation notation.

*a*

_{1}+

*a*

_{1}⋅

*r*+

*a*

_{1}⋅

*r*

^{2}+⋯+

*a*

_{1}⋅

*r*

^{(n-2)}+

*a*

_{1}⋅

*r*

^{(n-1)}=∑

^{n}

_{i=1}

*a*

_{1}⋅

*r*

^{(i-1)}.

When we sum a known number of terms in a geometric sequence, we get a finite geometric series. We can write out each term of a geometric sequence in the general form:

*a*=

_{n}*a*

_{1}⋅

*r*

^{(n-1)}

where

•

*n*is the index of the sequence;

•

*a*is the nth-term of the sequence;

_{n}•

*a*

_{1}is the first term;

•

*r*is the common ratio (the ratio of any term to the previous term).

By simply adding together the first

*n*terms, we are actually writing out the series

*S*=

_{n}*a*

_{1}+

*a*

_{1}⋅

*r*+

*a*

_{1}⋅

*r*

^{2}+⋯+

*a*

_{1}⋅

*r*

^{(n-2)}+

*a*

_{1}⋅

*r*

^{(n-1)}_ (i)

We may multiply the above equation by

*r*on both sides, giving us

*r⋅S*=

_{n}*a*

_{1}⋅

*r*+

*a*

_{1}⋅

*r*

^{2}+

*a*

_{1}⋅

*r*

^{3}+⋯+

*a*

_{1}⋅

*r*

^{(n-1)}+

*a*

_{1}⋅

*r*_ (ii)

^{n}You may notice that all the terms on the right side of (i) and (ii) are the same, except the first and last terms. If we subtract (i) from (ii), we are left with just

*r⋅S*–

_{n}*S*=

_{n}*a*

_{1}⋅

*r*–

^{n}*a*

_{1}

*S*(

_{n}*r*-1)=

*a*

_{1}(

*r*-1)

^{n}Dividing by (

*r*-1) on both sides, we arrive at the general form of a geometric series:

The second representation is derived by multiplying the first by (-1/-1). The former is more convenient where {

*r*>1} and the latter is more convenient where {

*r*<1}. Combined with (iii), we can see that:

**💎 Examples for which a ratio exists in {**

*r*<1}📌 Example 1: **The sum of a finite geometric series**

Find the sum of the series

^{12}

_{i=1}3(-2)

^{(i-1)}

✍ Solution:

This series is geometric with first term 3, ratio -2, and

*n*=12. We use the formula for the sum of the first 12 terms of a geometric series:

📌 Ex2. Find the sum of the first 10 terms of the geometric series: 4+2+1+⋯.

✍ Solution:

Method 1. Use the geometric series summation formula.

Write a formula:

List the information:

*u*

_{1}=4,

*r*=½,

*n*=10.

*S*

_{10}=?

Substitute and solve:

📌 Ex3. Find each sum. While it’s possible to determine what sort of series it is (and thus what formula I’ll need) by looking at the *k*th term formula, I suspect it will help if I write out a few terms of each one to convince you (and me) that we’ve got the right thing.

📌 Ex3a) ∑^{5}_{n=1} (-3)^{(n-1)}.

✍ Solution:

Find *n*, *a*_{1}, and *r*.

*n*=5-1+1=5

*a*

_{1}=(-3)

^{(1-1)}=1

*r*=-3

Substitute

*n*=5,

*a*

_{1}=1, and

*r*=-3 into the formula for the sum of a finite geometric series.

📌 Ex3b) First 7 terms of ∑^{∞}_{i=1} 8⋅(0.2)^{(i-1)}.

📌 Ex3c)

^{15}

_{k=1}50(0.8)

^{(k-1)}=50(0.8)

^{(1-1)}+50(0.8)

^{(2-1)}

+50(0.8)

^{(3-1)}+⋯+50(0.8)

^{(15-1)}

=50+50(0.8)+50(0.8)

^{2}+⋯+50(0.8)

^{14}.

I didn’t work out any of the terms any further because leaving them this way makes the pattern clearer (and takes less work). The series is geometric, the first term is 50, and the common ratio is 0.8. We need the sum of

*n*=14 terms.

**💎 Examples for which a ratio exists in {**

*r*>1}📌 Ex4: **Evaluate a Sum Written in Sigma Notation**

Evaluate ∑^{6}_{n=1} 5⋅2^{(n-1)}.

**Method 1**

Find the terms by replacing *n* with 1, 2, 3, 4, 5, and 6. Then add.

^{6}

_{n=1}5⋅2

^{(n-1)}=5⋅2

^{0}+5⋅2

^{1}+5⋅2

^{2}+5⋅2

^{3}+5⋅2

^{4}+5⋅2

^{5}

=5(1)+5(2)+5(4)+5(8)+5(16)+5(32)

=5+10+20+40+80+160=315

**Method 2**

Since the sum is a geometric series, you can use the formula

The sum of the series is 315.

📌 Ex5. Find the sum of the geometric series, ∑^{4}_{i=1} (2⋅4^{i}).

✍ Solution:

Method I. Draw the diagram: ∑^{4}_{i=1} (2⋅4^{i})

^{1}+2⋅4

^{2}+2⋅4

^{3}+2⋅4

^{4}

=8+32+128+512

Simply use your calculator and add the terms. Hence, the sum =680.

Method II may serve as a hint for the development of the formula to find the sum of a geometric series.

Method II. Finding the sum using a method without the use of a formula.

Let *S* be the sum, *S*_{4}=8+32+128+512

📌 Ex6: **Using S_{n} to Evaluate a Summation**

Find the following sum: ∑

^{10}

_{i=1}6⋅2

^{i}

✍ Solution:

Let’s write out a few terms in the sum.

^{10}

_{i=1}6⋅2

^{i}=6⋅2+6⋅2

^{2}+6⋅2

^{3}+⋯+6⋅2

^{10}

Do you see that each term after the first is obtained by multiplying the preceding term by 2? To find the sum of the 10 terms (

*n*=10), we need to know the first term,

*a*

_{1}, and the common ratio,

*r*. The first term is 6⋅2 or 12:

*a*

_{1}=12. The common ratio is 2.

Use the formula for the sum of the first

*n*terms of a geometric sequence.

*a*

_{1}(the first term) =12,

*r*=2, and

*n*=10 because we are adding ten terms.

Use a calculator.

^{10}

_{i=1}6⋅2

^{i}=12,276

📌 Ex7. Evaluate the sum of each series. You will need the first term *a*_{1}, the last term *a _{n}*, and

*n*.

📌 Ex7a. ∑

^{10}

_{k=1}5⋅3

^{(k-1)}

📌 Ex7b. ∑^{6}_{n=1} 5⋅2^{(n-1)}

✍ Solution:

Find *n*, *a*_{1}, and *r*.

*n*=6-1+1=6

*a*

_{1}=5⋅2

^{(1-1)}=5

*r*=2

Substitute

*n*=6,

*a*

_{1}=5, and

*r*=2 into the formula for the sum of a finite geometric series.

📌 Ex7c. ∑^{5}_{n=1} -4⋅3^{(n-1)}

✍ Solution:

Find *n*, *a*_{1}, and *r*.

*n*=5-1+1=5

*a*

_{1}=-4⋅3

^{(1-1)}=-4

*r*=3

Substitute

*n*=5,

*a*

_{1}=-4, and

*r*=3 into the formula for the sum of a finite geometric series.

📌 Ex7d. ∑^{6}_{n=1} 2⋅(1.4)〗^{(n-1)}

✍ Solution:

Find *n*, *a*_{1}, and *r*.

*n*=6-1+1=6

*a*

_{1}=2⋅(1.4)〗

^{(1-1)}=2

*r*=1.4

Substitute

*n*=6,

*a*

_{1}=2, and

*r*=1.4 into the formula for the sum of a finite geometric series.

📌 Ex8. What is the Value of *m* for which ∑^{m}_{k=1} 5⋅3^{k-1)}=65?

✍ Solution:

This is a geometric series because 5⋅3^{(k-1)} has the form *a⋅r*^{(k-1)},

*a*

_{1}=5⋅3

^{(1-1)}=5,

*a*

_{2}=5⋅3

^{(2-1)}=15,

*a*

_{3}=5⋅3

^{(3-1)}=45,

*a*=5,

*r*=3,

*n*=

*m*and

*S*=65.

_{m}Multiply through by 2.

^{m}-5

135=5⋅3

^{m}

Divide through by 5.

^{m}

Write 27 as a power of 3. Bases are the same, so the powers are equal.

^{3}=3

^{m}

m=3.

📌 Ex9. Find the value of *m* if: ∑^{m}_{k=1} 3⋅2^{(k-1)}=93

✍ Solution: