Problem:
20 + 20 1 5 + 20 2 5 + ⋯ + 40 2 0 + 2 0 5 1 + 2 0 5 2 + ⋯ + 4 0
Answer Choices:
A. 3000 3 0 0 0
B. 3030 3 0 3 0
C. 3150 3 1 5 0
D. 4100 4 1 0 0
E. 6000 6 0 0 0
Solution:
There are 101 1 0 1 101 ⋅ ( 20 + 40 2 ) = 101 ⋅ 30 = 3030 1 0 1 ⋅ ( 2 2 0 + 4 0 ) = 1 0 1 ⋅ 3 0 = 3 0 3 0
OR OR
There are 101 1 0 1
20 + 20 1 5 + 20 2 5 + ⋯ + 29 3 5 + 29 4 5 + 30 + 40 + 39 4 5 + 39 3 5 + ⋯ + 30 2 5 + 30 1 5 . 60 + 60 + 60 + ⋯ + 60 + 60 ⏟ 50 terms + 30 2 0 + 2 0 5 1 + 2 0 5 2 + ⋯ + 2 9 5 3 + 2 9 5 4 + 3 0 + 4 0 + 3 9 5 4 + 3 9 5 3 + ⋯ + 3 0 5 2 + 3 0 5 1 . 5 0 terms 6 0 + 6 0 + 6 0 + ⋯ + 6 0 + 6 0 + 3 0
Thus the sum is 50 ( 60 ) + 30 = 3030 5 0 ( 6 0 ) + 3 0 = 3 0 3 0
OR OR
Since 1 5 + 2 5 + 3 5 + 4 5 = 2 5 1 + 5 2 + 5 3 + 5 4 = 2
( 5 ( 20 ) + 2 ) + ( 5 ( 21 ) + 2 ) + ⋯ + ( 5 ( 39 ) + 2 ) + 40 = 5 ( 20 + 21 + ⋯ + 39 ) + 20 ( 2 ) + 40 = 5 ( 20 ⋅ 20 + 39 2 ) + 80 = 2950 + 80 = 3030. ( 5 ( 2 0 ) + 2 ) + ( 5 ( 2 1 ) + 2 ) + ⋯ + ( 5 ( 3 9 ) + 2 ) + 4 0 = 5 ( 2 0 + 2 1 + ⋯ + 3 9 ) + 2 0 ( 2 ) + 4 0 = 5 ( 2 0 ⋅ 2 2 0 + 3 9 ) + 8 0 = 2 9 5 0 + 8 0 = 3 0 3 0 .