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Determine the value of \(-1 + 2 + 3 + 4 - 5 - 6 - 7 - 8 - 9 + \dots + 10000\), where the signs change after each perfect square.

May 3, 2019

#1
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This reduces to the sum of alternating(+or-) cubes from 1 to the square root of 10,000 or 100 terms as follows:

sum(1 , 7 , 19 , 37 , 61 , 91 , 127 , 169 , 217 , 271 , 331 , 397 , 469 , 547 , 631 , 721 , 817 , 919 , 1027 , 1141 , 1261 , 1387 , 1519 , 1657 , 1801 , 1951 , 2107 , 2269 , 2437 , 2611 , 2791 , 2977 , 3169 , 3367 , 3571 , 3781 , 3997 , 4219 , 4447 , 4681 , 4921 , 5167 , 5419 , 5677 , 5941 , 6211 , 6487 , 6769 , 7057 , 7351 , 7651 , 7957 , 8269 , 8587 , 8911 , 9241 , 9577 , 9919 , 10267 , 10621 , 10981 , 11347 , 11719 , 12097 , 12481 , 12871 , 13267 , 13669 , 14077 , 14491 , 14911 , 15337 , 15769 , 16207 , 16651 , 17101 , 17557 , 18019 , 18487 , 18961 , 19441 , 19927 , 20419 , 20917 , 21421 , 21931 , 22447 , 22969 , 23497 , 24031 , 24571 , 25117 , 25669 , 26227 , 26791 , 27361 , 27937 , 28519 , 29107 , 29701)

=1,000,000 - which is the sum of the sequence.

May 3, 2019
#2
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Another simple way:

ALL the pluses and minues cancel each other out as follows:

-1 + 2 +3 + 4 - 5 -6 -7 -8 -9 +10 +11 + 12 + 13 + 14 + 15 + 16 - 17 - 18.........etc =64 - which is the 4^3 and the LAST cube in all these additions and subtractions. The process holds true for ALL 99 terms, leaving the value of the LAST cube, which is 100. Therefore:

100^3 =1,000,000 - which is the sum of this sequence.

May 4, 2019