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

= 6 + 7 + 9 + 13

= 35 .

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(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 .

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(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 .

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(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 .

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(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 .)

.

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

.

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(The summations must begin with i=1 in order to use the given formulas.)

= 10,497,600 - 2025 + 173,880 - 285

= 10,669,170 .

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(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 .

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= ( (-1) + 1 ) + ( (-1) + 1 ) + ... + ( (-1) + 1 ) + ( (-1) + 1 )

= 0 + 0 + ... + 0 + 0

= 0 .

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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.)

= (-1^{2} + 2^{2}) + (-2^{2} + 3^{2}) + (-3^{2} + 4^{2}) + (-4^{2} + 5^{2}) + ... + (-(*n*-1)^{2} + *n*^{2}) + (-*n*^{2} + (*n*+1)^{2})

(Group "like" terms.)

= -1^{2} + (2^{2} - 2^{2}) + (3^{2} - 3^{2}) + (4^{2} - 4^{2}) + (5^{2} - 5^{2}) + ... + ((*n*-1)^{2}-(*n*-1)^{2}) + (*n*^{2} - *n*^{2}) + (*n*+1)^{2}

= -1^{2} + (0) + (0) + (0) + (0) + ... + (0) + (0) + (*n*+1)^{2}

= (*n*+1)^{2} - 1

= *n*^{2} + 2*n* + 1 - 1

(*)

= *n*^{2} + 2*n* .

Second,

(**)

.

Equating expressions (*) and (**) we get that

,

,

or

.

This completes the proof.

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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.)

= (-1^{3} + 2^{3}) + (-2^{3} + 3^{3}) + (-3^{3} + 4^{3}) + (-4^{3} + 5^{3}) + ... + (-(*n*-1)^{3} + *n*^{3}) + (-*n*^{3} + (*n*+1)^{3})

(Group "like" terms.)

= -1^{3} + (2^{3} - 2^{3}) + (3^{3} - 3^{3}) + (4^{3} - 4^{3}) + (5^{3} - 5^{3}) + ... + ((*n*-1)^{3} - (*n*-1)^{3}) + (*n*^{3} - *n*^{3}) + (*n*+1)^{3}

= -1^{3} + (0) + (0) + (0) + (0) + ... + (0) + (0) + (*n*+1)^{3}

= (*n*+1)^{3} - 1

= *n*^{3} + 3*n*^{2} + 3*n* + 1 - 1

(*)

= *n*^{3} + 3*n*^{2} + 3*n* .

Second,

(**)

.

Equating expressions (*) and (**) we get that

,

,

,

or

.

This completes the proof.

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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 .

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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 .

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The formula

4*i* -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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