Express Some Sums in Expanded Form (Series)

Summation Notation

Sigma notation is used as a convenient shorthand notation for the summation of terms.
For example, we write


Here the symbol Σ (sigma) indicates a sum. The numbers at the top and bottom of sigma are called boundaries and tell us What numbers We substitute in to the expression for the terms in our sum. What comes after sigma is an algebraic expression representing terms in the sum. In the example above, n is a Variable and represents the terms in our sum.

Sigma notation

The symbol that is often used to express the concept of summation is the upper case Greek letter, sigma Σ.

The Σ notation is used in the following form: ∑ni=1 ui.
The notation is read as ‘the summation of all the ui’s from i=1 to i=n, where:
n is the number of terms as long as i=1
ui represents the terms that are being added
i is the variable that is used to increment to the next term.

Introduction To Sigma Notation

The Notation Itself

Sigma notation is a way of writing a sum of many terms, in a concise form. A sum in sigma notation looks something like this:


The Σ (sigma) indicates that a sum is being taken. The variable k is called the index of the sum. The numbers at the top and bottom of the Σ are called the upper and lower limits of the summation. In this case, the upper limit is 5, and the lower limit is 1. The notation means that we will take every integer value of k between 1 and 5 (so 1, 2, 3, 4, and 5) and plug them each into the summand formula (here that formula is 3k). Then those are all added together.

Sigma Notation (Summation Notation)
The Greek letter Σ is a mathematical symbol for finding the sum of the first n terms of a sequence ak.

The symbol Σ is the capital Greek letter sigma and is shorthand for ‘sum’. The lower and upper limits of the summation tells us which term to start with and which term to end with, respectively. For example, using the sequence an=2n-1 for n≥1, we can write the sum a3+a4+a5+a6 as


The index variable is considered a ‘dummy variable’ in the sense that it may be changed to any letter without affecting the value of the summation. For instance,

Some Examples

Example 1. Expand the series ∑7k=1 k2.

That’s it. Start with k=1 and substitute into the formula, increasing its value by 1 each time until you get to 7, the upper limit. The sigma sign indicates that it’s a sum. And since the question asks for expansion rather than a sum of the series, I’ll leave it like that.

Ex2. Evaluate ∑5n=2 n2.
Solution:
In this example we have used the letter n to represent the variable in the sum, rather than r. Any letter can be used, and we find the answer in the same Way as before:

Ex3. Evaluate ∑4r=1 (-1)r.
Solution:
Here, we need to remember that (-1)2=+1, (-1)3=-1, and so on. So

Ex4. Write out what is meant by the following:

The index Variable here is written as i instead of k. That’s ok. The most common Variables to use for indexes include i, j, k, m, and n.

Ex5. Evaluate ∑4r=1 r3.
Solution:
This is the sum of all the r3 terms from r=1 to r=4. So, we take each value of r, work out r3 in each case, and add the results. Therefore

Ex6. Write out explicitly what is meant by


Solution:
We must let k range from 1 to 5, cube each value of k, and add the results:

Ex7. Evaluate ∑5k=0 2k.
Solution:
Notice that, in this example, there are 6 terms in the sum, because we have k=0 for the first term:


What would we do if we were asked to evaluate

Now we know what this expression means, because it is the sum of all the terms 2k where k takes the values from 1 to n, and so it is

But we cannot give a numerical answer, as we do not know the value of the upper limit n.

Ex8. Evaluate ∑6r=1 ½r(r+1).
Solution:
You might recognise that each number ½r(r+1) is a triangular number, and so this example asks for the sum of the first six triangular numbers. We get

Ex9. Expand ∑5i=1 (2i+1)
Solution:

Ex10. Evaluate ∑5n=0 2n
Solution:

Ex11. Express the geometric series, ∑5i=1 (2⋅3(i-1)), as a sum of terms.
Solution:

Ex12. Expand the series indicated by the sigma notation.


Solution:
Expand the sigma notation.

Ex13: Find each sum. You will need to find each individual term and add them.
1. ∑5k=1 2⋅3(k-1)=2+6+18+54+162=242 (the sum)
Find each term: f(1)=2, f(2)=6, f(3)=18, f(4)=54, f(5)=162.

2. ∑4k=1 -7⋅2(k-1)=-7+(-14)+(-28)+(-56)=-105 (the sum)
Find each term: f(1)=-7, f(2)=-14, f(3)=-28, f(4)=-56

3. ∑7k=3 40⋅3(k-1)=360=1080+3240+9720+29160=43, 560

Ex14. Expand and evaluate:
a) ∑7k=1 (k+1) b) ∑5k=1 1/2k
solution:

Ex15: find the indicated sum
Ex15a) ∑4k=1 2k

=2⋅1+2⋅2+2⋅3+2⋅4
=2+4+6+8

Answer: 20

Ex15b) ∑6k=2 (2k-3)

=(2⋅2-3)+(2⋅3-3)+(2⋅4-3)+(2⋅5-3)
=1+3+5+7

Answer: 16

Ex15c) ∑4k=0 2k

=20+21+22+23+24
=1+2+4+8+16

Answer: 31

Ex15d) ∑8k=4 (-1)k

=(-1)4+(-1)5+(-1)6+(-1)7+(-1)8
=1+(-1)+1+(-1)+1

Answer: 1

Ex15f) ∑6k=1 5

=5+5+5+5+5+5+5

Answer: 30

Ex16. Write out what is meant by


Ex17. Evaluate ∑4k=1 k2.
Answers:
Ex16ans.

Ex17ans. 30.

Ex18. Write out what is meant by:

Ex19. Write out what is meant by the following:

Here, the index k takes the values 0, 1, 2, and 3. We’ll plug those each into 1/(k+1) and add them together.

Ex20. Evaluate


Solution:
Once again, we must remember how to deal with powers of -1:

One place you may encounter summation notation is in mathematical definitions. For example, summation notation allows us to define polynomials as functions of the form


for real numbers ak, k=0, 1, …, n.

Ex21. Express the sums ∑6i=1 i√2 and ∑7k=1 (-1)(k+1)/k2 in expanded notation.
Solution:

Ex22. Find the following sums.


Solution:
(a) We replace every occurrence of n with the values 0 through 4. We recall the factorials, n!=n⋅(n-1)⋅(n-2)⋅…⋅3⋅2⋅1 and get:

(c) We proceed as before, replacing the index n, but not the variable x, with the values 1 through 5 and adding the resulting terms.

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