Calculating the sum of each Arithmetic Series from its sigma notation

Finite Arithmetic Series
Remember that an arithmetic sequence is a set of numbers, such that the difference between any term and the previous term is a constant number, d, called the constant difference:

an=an+d(n-1)

where

n is the index of the sequence;

an is the nth-term of the sequence;

a1 is the first term;

d is the common difference.

When we sum a finite number of terms in an arithmetic sequence, we get a finite arithmetic series.

The simplest arithmetic sequence is when a1=1 and d=0 in the general form above; in other words all the terms in the sequence are 1:

ai=a1+d(i-1)
=1+0⋅(i-1)
=1
{ai}= {1; 1; 1; 1; 1; ⋯}

If we wish to sum this sequence from i=1 to any positive integer n, we would write

ni=1 ai=∑ni=1 1=1+1+1+⋯+1 _ (n times)

Since all the terms are equal to 1, it means that if we sum to n we will be adding n-number of 1’s together, which is simply equal to n:
ni=1 1=n

Another simple arithmetic sequence is when a1=1 and d=1, which is the sequence of positive integers:
ai=a1+d(i-1)
=1+1⋅(i-1)
=i
{ai}= {1; 2; 3; 4; 5; ⋯}

If we wish to sum this sequence from i=1 to any positive integer n, we would write
ni=1 i=1+2+3+⋯+n

This is an equation with a very important solution as it gives the answer to the sum of positive integers.

Interesting Fact
Mathematician, Karl Friedrich Gauss, discovered this proof when he was only 8 years old. His teacher had decided to give his class a problem which would distract them for the entire day by asking them to add all the numbers from 1 to 100. Young Karl realised how to do this almost instantaneously and shocked the teacher with the correct answer, 5050.

We first write Sn as a sum of terms in ascending order:

Sn=1+2+⋯+(n-1)+n _ (i)

We then write the same sum but with the terms in descending order:
Sn=n+(n-1)+⋯+2+1 _ (ii)

We then add corresponding pairs of terms from equations (i) and (ii), and we find that the sum for each pair is the same, (n+1):
2Sn=(n+1)+(n+1)+⋯+(n+1)+(n+1)

We then have n-number of (n+1)-terms, and by simplifying we arrive at the final result:
2Sn=n(n+1)
Snn(n+1)
Sn=∑ni=1 in(n+1)

Finite Arithmetic Sequences
The sum of a finite arithmetic sequence 1+2+⋯+n can be written in sigma notation as ∑ni=1 i, but that can alternatively be represented as ½n(n+1). So:

ni=1 in(n+1).

📌 Example 1. Use a formula to find 1+2+3+⋯+45
✍ Solution:
Use the formula ∑ni=1 i= ½n(n+1). In this application, it becomes

45i=1 i=½⋅45⋅46=1035.

Arithmetic Series
If we have an arithmetic sequence, adding up the terms gives us an arithmetic series. Once we realize it’s arithmetic, and we get the rule for each term in the form

an=a+(n-1)d.

Then there’s an easy formula for adding up the first n terms of an arithmetic series. It’s
Snn(a+an)

where Sn means the ‘sum of the first n terms’ (we could also use sigma notation, but this is briefer sometimes).

You’ll note that a is the first term of the series, and an is the last term. So our formula is pretty easy, it’s just taking the average of the first and last terms, and multiplying by the number of terms.

📌 Ex2. Find the sum of the series

25i=1 3i+4

Now this is the same rule as we added up in the previous example, but having 25 terms makes it very hard to do by hand. If it’s arithmetic, however we can use the formula to avoid having to write out the sequence in full.

Now the first few terms of the series are 7+10+13+⋯. So clearly it’s arithmetic, with a=7 and d=3 (d, remember, is the difference between successive pairs of terms.
Now the formula is

ai=a+(i-1)d
ai=7+(i-1)3=7+3i-3=3i+4

(which is the formula we started with!)

So now that we’ve shown it’s arithmetic, all we need to do to find the sum of the first 25 terms is

S25=½⋅25⋅(a+a25).

Now the first term is a=7, and the 25th term is
ai=3i+4
a25=3⋅25+4=79

so our formula is
S25=½⋅25⋅(7+79)=1075.

So the sum of the first 25 terms is 1075.

Equation 1. Sum of Arithmetic Sequence:
• The sum S of the first n terms of an arithmetic sequence ak=a+(k-1)d for k≥1 is

S=∑nk=1 akn(a1+an)

An application of the arithmetic sum formula which proves useful in Calculus results in formula for the sum of the first n natural numbers. The natural numbers themselves are a sequence(4) 1, 2, 3,…which is arithmetic with a=d=1. Applying Equation 1,

1+2+3+⋯+nn(n+1).

So, for example, the sum of the first 100 natural numbers is ½⋅100⋅101=5050.
_____
(4) This is the identity function on the natural numbers!

📌 Ex3: Find each sum.
📌 Ex3a. ∑40k=1 2k
✍ Solution:
I can write out the problem without summation notation as follows;

2(1)+2(2)+2(3)+⋯+2(40)
2+4+6+⋯+80.

This is asking me to add the first 40 terms of an arithmetic sequence. I will use the formula
Sn=½(a1+an).

In this case: S40=½⋅40⋅(2+80)=20(82)=1640
Answer: 1640

📌 Ex3b. ∑7n=1 (2n+1)
✍ Solution:

7n=1 (2n+1)=(2⋅1+1)+(2⋅2+1)+⋯+(2⋅7+1)
=3+5+⋯+15.

The first term of this series is 3 and the last term is 15. The number of terms is equal to the upper bound minus the lower bound plus one, which is 7-1+1 or 7. Therefore, a1=3, an=15, and n=7.
Find the sum of the series.
Snn(a1+an)
S7=½⋅7(3+15)
=63

📌 Ex3c. ∑7n=3 (3n+4)
✍ Solution:

7n=3 (3n+4)=(3⋅3+4)+(3⋅4+4)+⋯+(3⋅7+4)
=13+16+⋯+25.

The first term of this series is 13 and the last term is 25. The number of terms is equal to the upper bound minus the lower bound plus one, which is 7-3+1 or 5. Therefore, a1=13, an=25, and n=5. Find the sum of the series.
Snn(a1+an)
S(3 to 7)=½⋅5(13+25)
=95

📌 Ex3d. ∑20k=1 (60-5k)
✍ Solution:
I can write out the problem without summation notation as follows;

(60-5⋅1)+(60-5⋅2)+(60-5⋅3)+⋯+(60-5⋅20)
=55+50+45+⋯+(-40).

This is asking me to add the first 20 terms of an arithmetic sequence. I will use the formula
Sn=½(a1+an).

In this case: S20=½⋅20⋅(55+(-40))=10(15)=150
Answer: 150

📌 Ex3e.∑40k=1 (5-2k)
✍ Solution:
I can write out the problem without summation notation as follows;

(5-2⋅1)+(5-2⋅2)+(5-2⋅3)+⋯+(5-2⋅40)
=3+1+(-1)+⋯+(-75).

This is asking me to add the first 40 terms of an arithmetic sequence. I will use the formula
Sn=½(a1+an).

In this case:
S40=½⋅40⋅(3+(-75))=20(-72)=-1440.

Answer: -1440

📌 Ex3f. ∑60k=1 (2k-3)
✍ Solution:
I can write out the problem without summation notation as follows;

(2⋅1-3)+(2⋅2-3)+(2⋅3-3)+⋯+(2⋅60-3)-1+1+3+⋯+117.

This is asking me to add the first 60 terms of an arithmetic sequence. I will use the formula
Sn=½(a1+an).

In this case: S60=½⋅60⋅(-1+117)=30(116)=3480.
Answer: 3480

📌 Ex3g. ∑150n=1 (11+2n)
✍ Solution:

150n=1 (11+2n)=(11+2⋅1)+(11+2⋅2)+⋯+(11+2⋅150)
=13+15+⋯+311.

The first term of this series is 13 and the last term is 311. The number of terms is equal to the upper bound minus the lower bound plus one, which is 150-1+1 or 150. Therefore, a1=13, an=311, and n=150. Find the sum of the series.
Snn(a1+an)
S150=½⋅150⋅(13+311)
=24300

📌 Ex4. Find the sum of the arithmetic series, ∑20i=1 (3i-7).
✍ Solution:
Draw the diagram:

20i=1 (3i-7)=(3⋅1-7)+⋯+(3⋅20-7)
=-4+⋯+53.

Write a formula: Snn(u1+un)
List the information: u1=-4, n=20, u20=53.
S20=?

Substitute and solve: S20=½⋅20⋅(-4+53)=10⋅49=490.

Writing an arithmetic series using sigma notation involves using the nth term formula and leaving n and un as unknowns.

📌 Ex5. (Multiple Choice) Find ∑12k=1 (3k+9).
A 45 B 78 C 342 D 410
✍ Solution:
There are 12-1+1 or 12 terms, so n=12.

a1=3⋅1+9 or 12
a12=3⋅12+9 or 45.

Find the sum.
Sn=½⋅n⋅(a1+an)
S12=½⋅12⋅(12+45)
=342.

Option C is the correct answer.
Answer: C

📌 Ex6.
a. Write the sum of the first 15 terms of the arithmetic series 1+4+7+⋯ in sigma notation.
b. Find the sum.
✍ Solution
a. For the related arithmetic sequence, 1, 4, 7, ⋯ , the common difference is 3. Therefore,

an=1+(n-1)(3)
=3n-2.

The series is written as ∑15n=1 (3n-2). Answer

b. Use the formula Snn(2a1+(n-1)d) with a1=1, n=15, and d=3.

S15=½⋅15⋅[2⋅1+(15-1)⋅3]
S15=15⋅½⋅[2+42]
S15=15⋅22=330, Answer.

Note: Part b can also be solved by using the formula Snn(a1+an). First find

a15=a1+(n-1)d
=1+(15-1)⋅3
=1+43=43.
S15=½⋅15⋅(1+43)
S15=15⋅½⋅44
S15=330.

Notice that for this series there are 7½ pairs. The first 7 numbers are paired with the last 7 numbers, and each pair has a sum of 44. The middle number, 22, is paired with itself, making half of a pair.

📌 Ex7: Partial sums
For the sequence {an}={2n-1},
(a) evaluate the sum S25=∑25k=1 (2k-1) and
(b) find a formula for Sn.
✍ Solution:

Strategy: Let an=2n-1. To find S25 from Equation (3) requires a1, and a25, which the formula for an can provide. For (b), substitute 1 for a1 and 2n-1 for an in Equation (3) and simplify.

Follow the strategy.
(a) By Equation (3), S25=½⋅25⋅(a1+a25), find a1, and a25.
a1=2⋅1-1=1 and a25=2⋅25-1=49
Thus, S25=½⋅25⋅(1+49)=635.
(b) In general,
Snn(a1+an)=½n(1+(2n-1))=½n(2n)=n2
Hence, Sn=n2.

💎 General Formula for a Finite Arithmetic Series
If we wish to sum any arithmetic sequence, there is no need to work it out term-for-term. We will now determine the general formula to evaluate a finite arithmetic series. We start with the general formula for an arithmetic sequence and sum it from i=1 to any positive integer n:

ni=1 ai=∑ni=1 [ai+d(i-1)]
=∑ni=1 [ai+did]
=∑ni=1 [(aid)+di]
=∑ni=1 (aid) +∑ni=1 (di)
=∑ni=1 (aid) +dni=1 i
=(aid)ndn(n+1)
n(2ai-2d+dn+d)
n(2ai+dnd)
n(2ai+d(n-1)).

So, the general formula for determining an arithmetic series is given by
Sn=∑ni=1 [a1+d(i-1)]=½n[2ai+d(n-1)] _ (iii)

For example, if we wish to know the series S20 for the arithmetic sequence
ai=3+7(i-1), we could either calculate each term individually and sum them:
S20=∑20i=1 [3+7(i-1)]
=3+10+17+24+31+38+45+52
+59+66+73+80+87+94+101
+108+115+122+129+136
=1390

or, more sensibly, we could use equation (iii) noting that a1=3, d=7 and n=20 so that
S20=∑20i=1 [3+7(i-1)]
=½⋅20⋅[2⋅3+7(20-1)]
=1390.

This example demonstrates how useful equation (iii) is beneficial.

📌 Ex8. Evaluate ∑70n=4 (2n-4)
✍ Solution:
The question asks you to find the sum of the terms from n+4 to n+70 if the nth term is 2n-4.
Find the first term a.

a1=2(4)-4+4
a2=2(5)-4=6
a3=2(6)-4=8.

So the sequence is 4, 6, 8, and this is an arithmetic series.

To check d, calculate a2a1

d=a2a1=6-4=2
n=(70-4)+1=67.

There are 67 terms.
Now we can substitute these values into the formula to find the sum of 67 terms.
Snn[2a+(n-1)d]
S67=½⋅67⋅[2⋅4+(67-1)⋅2]
S67=33.5⋅(8 +132)=4690
70n=4 (2n-4)=4690.

Let’s read post Evaluating some Sums Written in Sigma Notation.