THE ALGEBRA OF SUMMATION NOTATION
The following problems involve the algebra (manipulation) of summation notation.
Summation notation is used to define the definite integral of a continuous function of
one variable on a closed interval. Let's first briefly define summation notation. If f(i)
represents some expression (function) involving i, then
meaning :
has the following
.
The "i=" part underneath the summation sign tells you which number to first plug into
the given expression. The number on top of the summation sign tells you the last
number to plug into the given expression. You always increase by one at each
successive step. For example,
= 3 + 6 + 11 + 18
= 38 .
We will need the following well-known summation rules.
1.
(n times) = cn, where c is a
constant.
2.
.
3.
.
4.
.
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Most of the following problems are average. A few are somewhat challenging.
If you are going to try these problems before looking at the solutions, you can
avoid common mistakes by using the formulas given above in exactly the form
that they are given. For instance, make sure that a summation begins with i=1
before using the above formulas.
SOLUTIONS TO THE ALGEBRA OF SUMMATION NOTATION
SOLUTION 1 :
= (5+1) + (5+2) + (5+4) + (5+8)
= 6 + 7 + 9 + 13
SOLUTION 2 :
(The above step is nothing more than changing the order and grouping of the original
summation.)
(Placing 3 in front of the second summation is simply factoring 3 from each term in
the summation. Now apply Rule 1 to the first summation and Rule 2 to the second
summation.)
2
= 400 + 15,150
= 15,550 .
SOLUTION 3 :
(Separate this summation into three separate summations.)
(Factor out the number 6 in the second summation.)
(Apply Rules 1, 2, and 3.)
= 2,686,700 - 120,600 + 1800
= 2,567,900 .
SOLUTION 4 :
3
(Since each summation begins with i=15, WE CANNOT USE THE RULES IN THE
FORM THAT THEY ARE GIVEN. Observe the following simple method to correct
this shortcoming.)
(Now apply Rules 1 and 2.)
= 4(11,325 - 105) + (136)
= 45,016 .
SOLUTION 5 :
(Note that
cancels, then
, then
, then ... all the way to
. Because of
this consecutive term cancellation, this type of summation is called a "telescoping"
sum. This cancellation will be shown in detail. First change the order of addition.)
4
(Now reassociate and collect "like" terms.)
(Recall that
.)
.
SOLUTION 6 :
(This is a "telescoping" sum. Group "like" terms and cancel.)
5
.
SOLUTION 7 :
(The summations must begin with i=1 in order to use the given formulas.)
= 10,497,600 - 2025 + 173,880 - 285
= 10,669,170 .
SOLUTION 8 :
6
(Recall that
if n is an integer.)
(Recall that
and
.)
= 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1) + 1 + (-1)
= (1 + (-1)) + (1 + (-1)) + (1 + (-1)) + (1 + (-1)) + (1 + (-1))
=0+0+0+0+0
=0.
SOLUTION 9 :
= ( (-1) + 1 ) + ( (-1) + 1 ) + ... + ( (-1) + 1 ) + ( (-1) + 1 )
= 0 + 0 + ... + 0 + 0
=0.
SOLUTION 10 :
7
There are several ways to prove that
. This proof uses a telescoping
sum. Consider the summation
ways. First, treat it as a telescoping sum. Then
. It can be evaluated in two different
(Commute the addition.)
= (-12 + 22) + (-22 + 32) + (-32 + 42) + (-42 + 52) + ... + (-(n-1)2 + n2) + (-n2 + (n+1)2)
(Group "like" terms.)
= -12 + (22 - 22) + (32 - 32) + (42 - 42) + (52 - 52) + ... + ((n-1)2-(n-1)2) + (n2 - n2) +
(n+1)2
= -12 + (0) + (0) + (0) + (0) + ... + (0) + (0) + (n+1)2
= (n+1)2 - 1
= n2 + 2n + 1 - 1
(*)
= n2 + 2n .
Second,
8
(**)
.
Equating expressions (*) and (**) we get that
,
,
or
.
This completes the proof.
SOLUTION 11 :
There are several ways to prove that
. This proof uses a
telescoping sum. Consider the summation
. It can be evaluated in
two different ways. First, treat it as a telescoping sum. Then
(Commute the addition.)
9
= (-13 + 23) + (-23 + 33) + (-33 + 43) + (-43 + 53) + ... + (-(n-1)3 + n3) + (-n3 + (n+1)3)
(Group "like" terms.)
= -13 + (23 - 23) + (33 - 33) + (43 - 43) + (53 - 53) + ... + ((n-1)3 - (n-1)3) + (n3 - n3) +
(n+1)3
= -13 + (0) + (0) + (0) + (0) + ... + (0) + (0) + (n+1)3
= (n+1)3 - 1
= n3 + 3n2 + 3n + 1 - 1
(*)
= n3 + 3n2 + 3n .
Second,
(**)
.
10
Equating expressions (*) and (**) we get that
,
,
,
or
.
This completes the proof.
SOLUTION 12 :
There is one nonobvious, but simple step in the solution of this problem. It requires
that you write a fraction as a sum or difference of partial fractions. For example,
11
is a partial fractions decomposition of
. Then a partial fraction decomposition of
is
so that
(This summation is a telescoping sum.)
(Now evaluate the limit.)
=1-0
=1.
SOLUTION 13 :
Note that in all of the following summations, letter i is a variable and letter n is a
constant (until the limit is evaluated). Then
12
(Now evaluate the limit.)
= 6 + (0) + 1 + (0) + (0)
=7.
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SOLUTION 14 :
The formula
4i -1 for i=1, 2, 3, 4, 5, ...
generates the given list of numbers. For example, the first number (i=1) in the list is
4(1)-1 = 3 .
The second number (i=2) in the list is
4(2)-1 = 7 .
The 30th number (i=30) in the list is
4(30)-1 = 119 .
The 120th number (i=12) in the list is
4(120)-1 = 479 .
Thus, the sum of the first 120 numbers in this list can now be computed as
= 29,040 - 120
= 28,920 .
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