Summation Notation Solutions

### SOLUTIONS TO THE ALGEBRA OF SUMMATION NOTATION

SOLUTION 1 :

= (5+1) + (5+2) + (5+4) + (5+8)

= 6 + 7 + 9 + 13

= 35 .

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.)

= 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 :

(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.)

(Now reassociate and collect "like" terms.)

(Recall that .)

.

SOLUTION 6 :

(This is a "telescoping" sum. Group "like" terms and cancel.)

.

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 :

(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 :

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

= (-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,

(**)

.

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

= (-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,

(**)

.

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,

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

(Now evaluate the limit.)

= 6 + (0) + 1 + (0) + (0)

= 7 .

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 .