Suppose T = 1 + 2 + 3 + ... + 209 . Including the sum, how many sums of two or more consecutive positive integers have value T?
Is 209 the last term? If so, the sum of the sequence =21,945.
21,945 has this many consecutive sequences:
1 - 10972 to 10973 = 21945
2 - 7314 to 7316 = 21945
3 - 4387 to 4391 = 21945
4 - 3655 to 3660 = 21945
5 - 3132 to 3138 = 21945
6 - 2190 to 2199 = 21945
7 - 1990 to 2000 = 21945
8 - 1561 to 1574 = 21945
9 - 1456 to 1470 = 21945
10 - 1146 to 1164 = 21945
11 - 1035 to 1055 = 21945
12 - 987 to 1008 = 21945
13 - 717 to 746 = 21945
14 - 649 to 681 = 21945
15 - 610 to 644 = 21945
16 - 559 to 596 = 21945
17 - 502 to 543 = 21945
18 - 372 to 426 = 21945
19 - 357 to 413 = 21945
20 - 300 to 365 = 21945
21 - 279 to 348 = 21945
22 - 247 to 323 = 21945
23 - 184 to 278 = 21945
24 - 157 to 261 = 21945
25 - 145 to 254 = 21945
26 - 136 to 249 = 21945
27 - 99 to 231 = 21945
28 - 66 to 219 = 21945
29 - 51 to 215 = 21945
30 - 21 to 210 = 21945
31 - 1 to 209 = 21945
Note: let me know if I misunderstood your question.
