Consider the number pattern at the right. The sum of the numbers 1–10 is 55. The sum of the numbers 11–20 is 155. The sum of the numbers 21–30 is 255

Question

                                                                                                                                        

a. What is the sum of the numbers 31–40?                     

b. What is the sum of the numbers 101–110?

c. What kind of reasoning did you use in parts (a) and (b)?

d. Following is the development of a formula for the sum of n consecutive integers.

S = x + ( x  +1) + (x +2)  + … + (y - 2)  + (y -1) + y                              The sum of n integers from x to y

+ S = y + (y -1) + (y -2) + …. + (x +2) + (x +1) + x                              The same sum in reverse order

2 S = (x +y) + (x +y) + (x +y) + …. + (x +y) + (x +y) + (x +y)                      add the equations.

2 S 5 n(x 1 y)                                                                                                       There are n terms of (x +y).

S = n(x + y)                                                                                                             Divide each side by 2.

              2

Use the formula to find the sum of the numbers 101–110.

e. What kind of reasoning did you use in part (d)?

Answer 1

31 to 40 = 355

101 to 110 = 1055

1 to 10 = 55

11 to 20 = 155

21 to 30 = 255

31 to 40 = 355

41 to 50 = 455

51 to 60 = 555

61 to 70 = 655

71 to 80 = 755

81 to 90 = 855

91 to 100 = 955

101 to 110 = 1055

keep the 55 ending

increase the "pre-fix" of -55, to match the beginning digit of the 10 digits sum

sum of 101 to 110 = sum of 1 to 110 after subtracting sum of 1 to 100

or use Gaus' formula: sum = n(n+1)/2

sum of 1 to 110 = 110(111)/2 = 55(111) = 6105

sum of 1 to 100 = 100(101)/2 - 50(101) = 5050

sum of 101 to 110 = 6105 -5050 = 1055