SEQUENCES AND SERIES
SEQUENCE AND SERIES
A sequence is a list of items/objects which have been arranged in a sequential way.
Example: 2, 4, 6, 8,…,10
A series can be highly generalized as the sum of all the terms in a sequence. However, there has
to be a definite relationship between all the terms of the sequence.
Example: 2 + 4 + 6 + 8+…+10
ARITHMETIC SERIES
Formula of πΊπ :
π
πΊπ = (ππ + ππ )
π
π
πΊπ = [πππ + (π − π)π
]
π
Example: Find the indicated value of series.
a. π + π + π + π + β― πΊπ
π
π
π
b. (− ) + (− ) + (− ) + (−π) + β― πΊπ
π
π
π
Solution
First, look for a pattern to identify the other terms until the nth term. Then, express
the sum of these terms.
a. πΊπ = π + π + π + π + π + ππ + ππ + ππ + ππ = ππ
π
π
π
πππ
b. πΊπ = (− ) + (− ) + (− ) + (−π) + (−π) + (−π) + (−π) = −
π
π
π
π
Summation Notation
Some Useful Formulas
π(π+π)
∑ππ=π π = π + π + π + π + β― + π =
i.
π
π(π+π)(ππ+π)
ii.
∑ππ=π ππ = ππ + ππ + ππ + ππ + β― + ππ =
iii.
π(π+π)
∑ππ=π ππ = ππ + ππ + ππ + ππ + β― + ππ = [
]
π
iv.
∑ππ=π ππ = ππ + ππ + ππ + ππ + β― + ππ =
π
π
π(π+π)(ππ +πππ+π−π)
ππ
Example 5: Write each in expanded form. Then, find its value.
a. ∑ππ=π(−π)π ππ
b. ∑ππ=π(π − ππ)π
Solution
Replace i consecutively with the integers starting with the value of the lower index
up to the value of the upper index.
a. ∑ππ=π(−π)π ππ
= (−π)π π(π) + (−π)π π(π) + (−π)π π(π) + (−π)π π(π) + (−π)π π(π) + (−π)π π(π)
= −π + π − π + π − ππ + ππ
=π
Here are some basic properties of summation:
If k is a real number,
i.
ii.
iii.
iv.
v.
Basic Properties of Summation
∑ππ=π π = ππ
∑ππ=π πππ = π ∑ππ=π ππ
∑ππ=π(ππ + ππ ) = ∑ππ=π ππ + ∑ππ=π ππ
∑ππ=π(ππ − ππ ) = ∑ππ=π ππ − ∑ππ=π ππ
∑ππ=π(ππ − ππ−π ) = ππ − ππ
Example: Evaluate the indicated sums by using the properties of the summation notation.
a. ∑ππ
b. ∑ππ=π −πππ
π=π ππ
c. ∑ππ=π(ππ + π)
d. ∑ππ=π πππ (ππ − π)
Solutions
a. ∑ππ
π=π ππ = (ππ)(ππ) = πππ
b. ∑ππ=π −πππ = −π ∑ππ=π ππ
= −π[(π)π + (π)π + (π)π + (π)π + (π)π ]
= −π[π + π + ππ + ππ + πππ]
= −π[πππ]
= −ππππ
b. ∑ππ=π(ππ + π) = ∑ππ=π ππ + ∑ππ=π π
= π ( π ) + π ( π ) + π ( π ) + π ( π ) + π ( π ) + π ( π ) + π (π ) + π
= π + π + ππ + ππ + ππ + ππ + ππ + ππ
= πππ
c. ∑ππ=π πππ (ππ − π) = π(π)π [π(π) − π] + π(π)π [π(π) − π] + π(π)π [π(π) − π] +
π(π)π [π(π) − π] + π(π)π [π(π) − π] + π(π)π [π(π) − π] +
π(π)π [π(π) − π]
= π + ππ + πππ + πππ + πππ + ππππ + ππππ
= ππππ
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