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Transcript

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. . 1 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. 13 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 . 14 15