#$&* course mth 164 12/12 7pmThank you for your concern the surgery went well. I am trying to get the fogginess out of my head from the pain meds and hope to take test 4 on Friday. If I don't make a 70 on it is there a chance I could try again before my final grade. SOLUTIONS/COMMENTARY ON QUERY 16
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (-1)^(1-1)*1/(2*1-1) = 1 (-1)^(2-1)*2/(2*2-1) = (-1)^1*2/3 = -.667 or 2/3 (-1)^(3-1)*3/(3*2-1) = (-1)^2*3/5 = .6 or 3/5 (-1)^(4-1)*4/(4*2-1) = (-1)^3*4/7 = -.5714 or -4/7 (-1)^(5-1)*5/(5*2-1) = (-1)^4*5/9 = .5556 or 5/9 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Subsituting we obtain a(1) = (-1)^(1-1) * 1 / (2*1 - 1) = (-1)^0 * 1 / 1 = 1 a(2) = (-1)^(2-1) * 2 / (2*2 - 1) = (-1)^1 * 2 / 3 = -2/3 a(3) = (-1)^(3-1) * 3 / (2*3 - 1) = (-1)^2 * 3 / 5 = 3/5 a(4) = (-1)^(4-1) * 4 / (2*4 - 1) = (-1)^3 * 4 / 7 = -4/7 a(5) = (-1)^(5-1) * 5 / (2*5 - 1) = (-1)^4 * 5 / 9 = 5/9 **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: Query 11.1.16 nth term of 2/3,4/9,8/27, 16/81,...
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (2/3)^n confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The nth term of 2/3,4/9,8/27, 16/81,... is a(n) = { (2^n) / (3^n) } ** Good. Note that (2^n) / (3^n) = (2/3)^n. The sequence consists of powers of 2/3. We can therefore write the sequence as a(n) = (2/3)^n. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.1.18 nth term of 1, 1/2, 3/, 1/4, ...
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Couldn’t answer this question because I don’t know what 3/? is. There is a number missing in the question. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: a sub n= 1/n if n is even a sub n = n if n is odd ** Good, but the even-odd thing requires a decision on the part of the user. Better to find a single expression to define the sequence. If you raise n to the +1 power when n is odd and to the -1 power when n is even, you get the correct sequence. So we want to find an expression that's 1 whenever n is odd and -1 whenever n is even. Noting that the powers of -1 alternate between 1 and -1, we could try (-1)^n. However that gives -1 when n is odd and 1 when n is even. So we try (-1)^(n-1), which does give us 1 when n is odd and -1 when n is even. So we use this expression for our exponent, and get the single expression that defines the sequence: n^( -1^(n-1)). **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.1.30 a1 = -1, a2 = 1, a(n+2) = a(n+1) + n an.
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am confused by this question. How do you know when to use a1 or a2, the solution does not even use a1 that I can see. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** For n = 1 we get a(1+2) = a(1+1) + 1 * a(1), or a(3) = a(2) + 1 * a(1). This gives us a(3) = 1 + 1 * -1 = 1 - 1 = 0. Then a(4) = a(3) + 2 a(2) = 0 + 2 * 1 = 2. Then a(5) = a(4) + 3 a(3) = 2 + 3 * 0 = 2. Then a(6) = a(5) + 4 a(4) = 2 + 4 * 2 = 10. **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:
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14:37:46 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3(1)-7 = -4 3(2)-7 = -1 3(3)-7 = 2 3(4)-7 = 5 3(5)-7 = 8 3(6)-7 = 11 The total sum is 21 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** We are to sum the first six terms of the sequence. We calculate: a(1) = ((3(1)-7)=-4 a(2) = ((3)(2)-7=-1 a(3) = ((3)(3)-7)=2 a(4) = ((3)(4)-7)=5 a(5) = ((3)(5)-7)=8 a(6) = ((3)(6)-7)=11 Our total is a(1) + a(2) + a(3) + a(4) + a(5) + a(6) = 21. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.1.50 write out sum( (k+1)^2, k, 1, n).
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YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1+1)^2 = 4 (2+1)^2 = 9 (3+1)^2 = 16 (4+1)^2 = 25 (5+1)^2 = 36 (6+1)^2 = 49 The sum is 139 I am not sure this is what the question is asking for. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Our terms are a(1) = (1+1)^2=4 a(2) = (2+1)^2=9 a(3) = (3+1)^2=16 a(4) = (4+1)^2=25 a(5) = (5+1)^2=36 . . . a(n) = (n+1)^2 The sum is therefore written as 4+9+16+25+36+49...+ (n+1)^2 **
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.1.60 express using summation notation 1 + 3 + ... + [2(12)-1]
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14:44:02 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: On top of the sigma would be 12, under the sigma would be k = 1, to the right of sigma would be 2k-1. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: *&*& Using summation notation . we would have at the top of the sigma the letter 12 to represent how many numbers the sequence goes through. at the bottom of sigma we have k=1 and to the right of it the equation 2k-1 In DERIVE notation the expression is sum(2k-1, k, 1, 12), indicating the sum of the expression 2k-1 from k = 1 to k = 12. *&*&
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.1.72 (5th ed #68) un = [ (1 + `sqrt(5))^n - (1 - `sqrt(5)^n ] / (2^n `sqrt(5) )
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14:50:43 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t understand how to answer this question. confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: STUDENT SOLUTION for part a) we can say that u sub 1 equals 1 because in the Fibonacci sequence u sub 1 and u sub 2 are equal. and the third term of the sequence is the sum of u sub 1 and u sub 2. b) in the sequence each new number is the addition of the current number and the starting number which is in this case u sub n. so u sub n+1 + u sub n= u sub n+2 c)we know that u sub n is the Fibonacci sequence for various reasons. the main reason is the fact that if both equations in the numerator are the same and the exponents are the same then those equations will according to the sequence will be an even number which can be divided by the denominator.
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14:50:43
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Explain how you showed that u1 = 1 and u2 = 1.
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20:23:48 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: *&*& For u1 you let n = 1 and for u2 you let n = 2: u1 = [ (1 + `sqrt(5))^1 - (1 - `sqrt(5)^1 ] / (2^1`sqrt(5) ) = ( 1 + 'sqrt(5) -1 + 'sqrt(5) ) / 2 'sqrt(5) = ( 2 'sqrt(5) ) / ( 2 'sqrt(5) ) = 1 and u2 = [ (1 + `sqrt(5))^2 - (1 - `sqrt(5)^2 ] / (2^2`sqrt(5) ) = ( 1 + 'sqrt(5) + 'sqrt(5) + 5 ) - ( 1 -'sqrt(5) -'sqrt(5) + 5 ) / 4 'sqrt(5) = ( 4 'sqrt(5) ) / ( 4 'sqrt(5) ) = 1 *&*&
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20:23:49
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Explain how you showed that u(n+2) = u(n+1) + un.
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20:28:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: After workin out u1 and u2, which equalled 1 and u3 , which equalled 2. I then aworked u3 which is u2 plus u1 to get 1 + 1 = 2 u4 = u3 + u2 = 2 + 1 = 3 Because obtaineing the third term required the 1st and 2nd terms ( and so forth) this is the Fibonacci sequence. ** To show this for general n you have to simplify the expression for u(n+1) + un to show that it's the same as that for u(n+2). [ (1 + `sqrt(5))^(n+2) - (1 - `sqrt(5))^(n+2) ] / (2^(n+2) `sqrt(5) ) = [ (1 + `sqrt(5))^(n+1) - (1 - `sqrt(5))^(n+1) ] / (2^(n+1) `sqrt(5) ) + [ (1 + `sqrt(5))^n - (1 - `sqrt(5)^n ] / (2^n `sqrt(5) ). You begin by multiplying both sides 2^n * `sqrt(5), which eliminates part of the denominator. They multiply both sides by the common denominator 4 of the remaining expressions. Then use the distributive law, being careful, expanding everything into individual terms then show that everything just cancels out, leaving you 0 = 0.** for the Fibonacci sequence we must be able to say u sub 3= u sub 1 + u sub 2 which in this case would equal 2. then we can move up to u sub 4 which would be u sub 2+ u sub 3 which = 3 and u sub 5= u sub 3 + u sub 4= 5 and this equation fits that pattern. ** You obtained your u1, u2, u3 ... values by referring to the Fibonacci sequence. Then you used this to justify your statement that the expression represents the Fibonacci sequence. So you are saying that the sequence represents the Fib seq because the expression represents the Fib seq. That's circular. If you had algebraically evaluated the expression for n = 1, n = 2, etc., and shown that the numbers are indeed identical to those of the Fibonacci sequence, then you wouldn't have proven the result but you would have a good indication that it works. **
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14:51:32
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.2.18 nth term of arithmetic sequence with a = 0, d = `pi. **** What is your expression for the nth term of the sequence? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (n-1)*pi + 0 = pi(n) - pi confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: NOT-QUITE-CORRECT STUDENT SOLUTION a sub n= pi(n)-n (pi)(n)-pi ** Term n+1 would then be `pi * (n+1) - (n+1), and the difference between the nth term and the n+1 term would be `pi * (n+1) - (n+1) - [ `pi * n - n] = `pi * n+ `pi * 1 - `pi * n - n - 1 + n = `pi - 1. But the difference should be `pi. The correct expression for the nth term is a(n) = `pi * n. ** The sequence begins with a(0) = 0 and each term adds pi to the preceding term. So a(1) = a(0) + pi = 0 + pi = pi a(2) = a(1) + pi = pi + pi = 2 pi a(3) = a(2) + pi = 2 pi + pi = 3 pi etc. At the nth step we get a(n) = n pi.
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.2.24 7th term of 2 `sqrt(5), 4 `sqrt(5), 6 `sqrt(5), ...
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14:53:01 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Not-quite-correct student solution: a sub 7= 14(sq.rt.5) a7 = 2 'sqrt(5) + 12 Here's how I got it. I used a(n) = a + (n-1)d ; a = 2 'sqrt(5) ; d = 2 a7 = 2 'sqrt (5) + (7-1) 2 = 2 'sqrt (5) + 12 ** If the sequence is as I stated above, the common difference is 2 `sqrt(5) and the 7th term is 14 `sqrt(5). **
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14:54:20
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.2.30 5th term -2, 13th term 30. ......!!!!!!!!................................... 14:58:53 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A1 = -18 and d= 4 (n-1)*4-18 = 4n-4-18 = 4n-22 The nth term is 4n - 22 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If a(n) = a + n * d then we have a(5) = a + 5 * d = -2 and a(13) = a + 13 * d = 30 . Subtracting a + 5 * d = -2 from a + 13 * d = 30 we get 8 * d = 32 so that d = 4. Substituting this into a + 5 * d = -2 we get a + 5 * 4 = -2 so that a = -2 - 5 * 4 = -22. Our expression for a(n) is therefore a(n) = -22 + 4 * n. we can find the value for d and the first number in the sequence by saying a sub 5=a +4d=-2 a sub 13=a +12d=30 by subtracting the second equation from the first we get -8d=-32 so d=4 we can then find the value of a sub 1 by plugging in the value of d=4 to the equation a sub 5. so a= -2- 4d so a=-2-(4)(4) so a sub 1=-18 the formula we could use is a sub n=a +(4n-4) ** If d = 4 and a(1) = -18 then we have a(n) = a(n-1) + 4, with a(1) = -18. **
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14:58:57
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&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.2.36 sum of -1+3+7+...+(4n-5).
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15:03:32 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: n/2(-1 + (4n-5) = n/2(4n-6) confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: by using the equation S sub n = n/2(a+a sub n) we get s sub n= n/2(-1+(4n-5)= (n/2)(4n-6) ** The sum of -1 and (4n-5) is 4n - 6. If we pair the numbers in the usual fashion this is the sum of every pair. We have n/2 pairs so the sum is n/2 ( 4n - 6) = 2 n^2 - 3 n. ** ** Good. This simplifies to give us n * ( 2n - 3 ). If expanded form is desired you would have 2 n^2 - 3 n. ** Using the formula Sn = n/2 (a + a(n) ) I got... Sn = n /2 ( 8 + (4n - 5 )) = n/2(4n +3) ** Shouldn't that be Sn = n/2 ( a(1) + a(n) ) = n/2 (-1 + (4n-5)) = n/2 (4n - 6) = 2 n^2 - 3 n? **
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15:03:40 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: **** Query 11.2.52 staircase 30 steps, bottom requires 100 bricks, each successive step 2 less than previous.
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15:10:58 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 102-2*n, 102-2(30) = 42, There are 42 bricks on the 30th step If you add all the steps together you get 2130 bricks. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Step n requires a0 - 2 n bricks. Since step 1 requires 100 bricks, a1 = a0 - 2 * 1 = 100. Solving for a0 we get a0 = 102. So step n requires 102 - 2 * n bricks. That is, an = 102 - 2 * n. It follows that a30 = 102 - 2 * 30 = 42. The total number of brick is number of pairs * total for each pair = n/2 ( a1 + an) = 30/2 ( 100 + 42) = 15 * 142 = 2130. **
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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: **** Query 11.2.52 staircase 30 steps, bottom requires 100 bricks, each successive step 2 less than previous.
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15:10:58 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 102-2*n, 102-2(30) = 42, There are 42 bricks on the 30th step If you add all the steps together you get 2130 bricks. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Step n requires a0 - 2 n bricks. Since step 1 requires 100 bricks, a1 = a0 - 2 * 1 = 100. Solving for a0 we get a0 = 102. So step n requires 102 - 2 * n bricks. That is, an = 102 - 2 * n. It follows that a30 = 102 - 2 * 30 = 42. The total number of brick is number of pairs * total for each pair = n/2 ( a1 + an) = 30/2 ( 100 + 42) = 15 * 142 = 2130. **
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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!