12.5 Sigma Notation and the nth term

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*
By the end of the section students will be able to expand a summation given in
sigma notation, determine the sum of an arithmetic series using sigma notation
and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
Assignment #45
No book assignment,
instead it is a
worksheet.
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
* Capitol Greek letter sigma Σ
* Sigma represents a Summation
* The bottom it the start value, lower bound of summation
* The top is the end value, upper bound of summation
* The letter used in the lower bound of summation is called
the index of summation
πŸ”
π‘Žπ’Œ =
π’Œ=𝟏
π‘ŽπŸ
π’‘π’π’–π’ˆ π’Šπ’
π’π’π’˜π’†π’“
𝒃𝒐𝒖𝒏𝒅
+ π‘Ž2 + π‘Ž3 + π‘Ž4 + π‘Ž 5 +
π‘ŽπŸ”
π’‘π’π’–π’ˆ π’Šπ’
𝒖𝒑𝒑𝒆𝒓
𝒃𝒐𝒖𝒏𝒅
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
A.
8
3π‘˜ + 7 =
π‘˜=2
13 + 16 + 19 + 22 + 25 + 28 + 31
π‘˜=2
π‘˜=3
π‘˜=4
π‘˜=5
π‘˜=6
π‘˜=7
π‘˜=8
This is an arithmetic series with common difference +3
B.
8
𝑖=4
1
2
3
𝑖−3
=
2
2
2
2
2
+
+
+
+
3
9
27 81 243
𝑖=4
𝑖=5
𝑖=6
𝑖=7
This is a geometric series with common
𝑖=8
1
ratio 3
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
C.
5
−2π‘ž + 1 =
π‘ž=1
−1 − 3 − 5 − 7 − 9
This is an arithmetic series with common difference -2
D.
4
5 2
π‘˜+1
=
π‘˜=1
20 + 40 + 80 + 160
This is a geometric series with common ratio 2
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
* Arithmetic Series
* Geometric Series
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
π‘Ž1 1 − π‘Ÿ 𝑛
𝑆𝑛 =
1−π‘Ÿ
* Infinite Geometric series
π‘Ž1
𝑆=
1−π‘Ÿ
For our purposes you will only need to use the ARITHMETIC
for sigma notation problems.
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
B.
A.
13
9
7π‘˜ − 1 =
π‘˜=1
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
𝑛=9
π‘Ž1 = 7 1 − 1 = 6
π‘Ž9 = 7 9 − 1 = 62
9
𝑆9 = 6 + 62
2
𝑆9 = 9 34 = 306
1
π‘˜ + 10 =
2
π‘˜=1
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
𝑛 = 12
1
21
π‘Ž1 = 1 + 10 =
2
2
1
33
π‘Ž13 = 13 + 10 =
2
2
13 21 33
13 54
𝑆13 =
+
=
2 2
2
2 2
351
=
2
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
C.
D.
13
25
10π‘˜ − 25 =
−3π‘˜ + 17 =
π‘˜=1
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
𝑛 = 25
π‘Ž1 = −3 1 + 17 = 14
π‘Ž25 = −3 25 + 17 = −58
25
𝑆25 =
14 − 58
2
𝑆25 = 25 −22 = −550
π‘˜=1
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
𝑛 = 13
π‘Ž1 = 10 1 − 25 = −15
π‘Ž13 = 10 13 − 25 = 105
13
𝑆13 =
−15 + 105
2
𝑆13 = 13 45 = 585
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
* Move all terms to one side so that one side is zero
* A quadratic has TWO solutions that can be found by…
* X-box Factoring
* Guess and Check factoring
* Quadratic formula
* Note: for Series
* Do we have fractional terms? (e.g. first term,
* Do we have negative terms? (e.g. -4th term?)
1π‘‘β„Ž
1
2
term?)
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
A.
𝑛
2π‘˜ − 12 = −πŸπŸ–
π‘˜=1
𝑛
𝑺𝒏 = π’‚πŸ + 𝒂𝒏
2
𝑛
−πŸπŸ– = −𝟏𝟎 + πŸπ’ − 𝟏𝟐
2
𝑛
−18 = 2𝑛 − 22
2
−18 = 𝑛 𝑛 − 11
𝑺𝒏 = −πŸπŸ–
−18 = 𝑛2 − 11𝑛
π’‚πŸ = 2 1 − 12 = −𝟏𝟎
0 = 𝑛2 − 11𝑛 + 18
𝒂𝒏 = πŸπ’ − 𝟏𝟐
0 = 𝑛 − 2 (𝑛 − 9)
𝑛
−πŸπŸ– = −𝟏𝟎 + πŸπ’ − 𝟏𝟐
2
𝑛 = 2, 9
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
B.
𝑛
−2π‘˜ − 8 = −πŸ‘πŸ”
𝑛
−πŸ‘πŸ” = −𝟏𝟎 − πŸπ’ − πŸ–
2
−36 =
π‘˜=1
𝑛
𝑺𝒏 = π’‚πŸ + 𝒂𝒏
2
𝑺𝒏 = −πŸ‘πŸ”
π’‚πŸ = −2 1 − 8 = −𝟏𝟎
𝒂𝒏 = −πŸπ’ − πŸ–
𝑛
−πŸ‘πŸ” = −𝟏𝟎 − πŸπ’ − πŸ–
2
𝑛
−2𝑛 − 18
2
−36 = 𝑛 −𝑛 − 9
−36 = −𝑛2 − 9𝑛
𝑛2 + 9𝑛 − 36 = 0
𝑛 − 3 𝑛 + 12
𝑛 = 3, −12
𝑛=3
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
C.
𝑛
(π‘˜ − 8) = −27
𝑛
−27 = 𝑛 − 15
2
−54 = 𝑛 𝑛 − 15
π‘˜=1
𝑆𝑛 =
𝑛
π‘Ž + π‘Žπ‘›
2 1
−54 = 𝑛2 − 15𝑛
𝑆𝑛 = −27
0 = 𝑛2 − 15𝑛 + 54
π‘Ž1 = 1 − 8 = −7
0= 𝑛−6 𝑛−9
π‘Žπ‘› = 𝑛 − 8
𝑛 = 6, 9
−27 =
𝑛
−7 + 𝑛 − 8
2
−27 =
𝑛
𝑛 − 15
2
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as
evidenced by completion of an exit slip.
D.
𝑛
−3π‘˜ + 9 = −21
π‘˜=1
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
𝑆𝑛 = −21
π‘Ž1 = −3 1 + 9 = 6
π‘Žπ‘› = −3𝑛 + 9
𝑛
−21 = 6 − 3𝑛 + 9
2
−21 =
𝑛
−3𝑛 + 15
2
𝑛
−21 = −3𝑛 + 15
2
−42 = 𝑛 −3𝑛 + 15
−42 = −3𝑛2 + 15𝑛
3𝑛2 − 15𝑛 − 42 = 0
3 𝑛2 − 5𝑛 − 14 = 0
3 𝑛−7 𝑛+2 =0
𝑛 = 7, −2
𝑛=7
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
1.
Find the number of terms (n) needed for the series
below to have a sum of 14
𝑛
−3π‘˜ + 14 = 14
π‘˜=1
By the end of the section students will be able to expand a summation given in sigma notation, determine the sum of an
arithmetic series using sigma notation and determine the number of terms in a arithmetic sequence for a given sum as evidenced
by completion of an exit slip.
1.
Find the number of terms (n) needed for the series below to
have a sum of 14
𝑛
−3π‘˜ + 14 = 14
π‘˜=1
𝑛
𝑆𝑛 = π‘Ž1 + π‘Žπ‘›
2
𝑛
14 = 11 + −3𝑛 + 14
2
28 = 𝑛 −3𝑛 + 25
0 = −3𝑛2 + 25𝑛 − 28
0 = − 3𝑛2 − 25𝑛 + 28
0 = − 3𝑛 − 4 𝑛 − 7
4
𝑛 = ,7
3
𝑛=7
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