Discrete Math Proof?
2016-12-01 12:16 am
I need help proving that ((k(2k-1)(2k+1)) / 3) + (2k+1)^2 is equal to (k+1(2k+1)(2k+2)) / 3
回答 (3)
2016-12-01 1:00 am
?
2016-12-01 12:32 am
Let k = 0
((k(2k-1)(2k+1)) / 3) + (2k+1)^2
= ((0(-1)(1)) / 3) + (1)^2
= 0 + 1
= 1
(k+1(2k+1)(2k+2)) / 3
= (0+1(1)(2)) / 3
= 2/3
Therefore, it not possible to prove that ((k(2k-1)(2k+1)) / 3) + (2k+1)^2 is equal to (k+1(2k+1)(2k+2)) / 3
2016-12-01 12:23 am
((k(2k-1)(2k+1)) / 3) + (2k+1)^2 =
(1/3)(2k+1)[k(2k-1) + 3(2k+1)] =
(1/3)(2k+1)[2k^2 + 5k + 3] =
(1/3)(2k+1)(k+1)(2k+3).
It does not equal what you propose.
(1/3)(2k+1)[k(2k-1) + 3(2k+1)] =
(1/3)(2k+1)[2k^2 + 5k + 3] =
(1/3)(2k+1)(k+1)(2k+3).
It does not equal what you propose.
收錄日期: 2021-04-23 23:54:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20161130161637AAuTAdw