Math 103 - (1) Compute the following sums: (a) (b) (c) (d)...

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(1) Compute the following sums: (a) 150 i =1 2 (b) 50 n =1 n (c) 60 n =10 n (d) 25 n =1 3 n 2 (e) 55 i =1 ( i +2) (f) 500 k =100 k (g) 50 k =2 ( k 2 - 2 k +1) (h) 20 m =10 m 3 Solution: For the solutions to these, we will use the following results, which have been established by Gauss’s formula, and the notation shown below for convenience: S 0 ( n )= n ± i =1 1= nS 1 ( n n ± i =1 i = n ( n 2 S 2 ( n n ± i =1 i 2 = n ( n +1)(2 n 6 S 3 ( n n ± i =1 i 3 = ² n ( n 2 ³ 2 (a) 150 i =1 2=2 150 i =1 1=2(150)=300 (b) 50 n =1 n = (50)(50+1) 2 =1275 (c) 60 n =10 n = 60 n =1 n - 9 n =1 n = (60)(61) 2 - (9)(10) 2 =1785 (d) 25 n =1 3 n 2 =3 25 n =1 (25)(26)(51) 6 =5525 (e) 55 i =1 ( i +2)= 55 i =1 i + 55 i =1 2= (55)(56) 2 +2(55)=1650 . 500 k =100 k = 500 k =1 k - 99 k =1 k = (500)(501) 2 - (99)(100) 2 =120300 (g) 50 k =2 ( k 2 - 2 k +1)= 50 k =2 k 2 - 2 50 k =2 k + 50 k =2 1 =( 50 k =1 k 2 - 1) - 2( 50 k =1 k - 1) + ( 50 k =1 1 - 1) (50)(51)(101) 6 - 1) - 2( (50)(51) 2 - 1) + (1(55) - 1) = 40430 (h) 20 m =10 m 3 = 20 m =1 m 3 - 9 m =1 m 3 =[ (20)(21) 2 ] 2 - [ (9)(10) 2 ] 2 =46125 (2) A clock at London’s Heathrow airport chimes every half hour. At the beginning of the n ’th hour, the clock chimes n times. (For example, at 8:00 am the clock chimes 8 times, at 2:00 pm the clock chimes fourteen times, and at midnight the clock chimes 24 times.) The clock also chimes once at half-past every hour. Determine how many times in total the clock chimes in one full day. Use sigma notation to write the form of the series before ±nding its sum. Solution: ¡¡ There is no solution for this problem. ¿¿ 1
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(3) Use the sigma summation notation to set up the following problems, and then apply known formulae to compute the sums. (a) Find the sum of the ±rst 100 even numbers. (b) Find the sum of the ±rst 100 odd numbers. (c) The sum of the ±rst 100 integers. Solution: (a) An even number can be written in the form 2 n . Thus the sum of the ±rst ±fty such numbers is 50 n =1 (2 n )=2 50 n =1 n =2 S 1 (50) = 2550. (b) An odd number can be written in the form 2 n - 1 where n =1 , 2 .. . Thus the sum of the ±rst ±fty such numbers is 50 n =1 (2 n - 1) = 2 50 n =1 n - 50 n =1 1=2 S 1 (50) - S 0 (50) = 2550 - 50 = 2500 (c) The sum of the ±rst 100 integers is 100 n =1 n = S 1 (100) = 5050 which is the same as the total of the sums in parts (a) and (b) (4) Consider all the integers that are of the form n ( n - 1) where n , 2 , 3 .. . Find the sum of the ±rst 50 such numbers. Solution: ¡¡ There is no solution for this problem. ¿¿ (5) Aseto fJapaneselacquerboxeshavebeenmadeto±tper fect lyoneins idetheother . (When objects ±t inside one another perfectly, we say that they form a ”nested” set.) All the boxes are cubical, and they have sides of lengths 1,2,3. .15 inches. Find the total volume enclosed by all the boxes combined. (You may assume that the walls are negligeably thin). Solution: ¡¡ There is no solution for this exercise. ¿¿ (6) Afram ingshopusesasquarep ieceofmattcardboardtocreateasetofsquareframes , one cut out from the other, with as little wasted as possible. The original piece of cardboard is 50 cm by 50 cm. Each of the ”nested” square frames (see Prob 5 for defn of nested) has a border 2 cm thick. How many frames in all can be made from this original piece of cardboard? What is the total area that can be enclosed by all these frames together?
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This note was uploaded on 01/12/2011 for the course MATH 103 taught by Professor Israel during the Spring '07 term at The University of British Columbia.

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Math 103 - (1) Compute the following sums: (a) (b) (c) (d)...

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