close
標題:
mathematical induction
發問:
最佳解答:
Let P(n): (x+y)^n - x^n - y^n is divisible by xy(x+y). When n = 1: (x + y) - x - y = 0 which is divisible by xy(x + y) Assume that P(k) is true, (x+y)^k - x^k - y^k is divisible by xy(x+y). When n = k + 1 (x + y)^(k + 1) - x^(k + 1) - y^(k + 1) = (x + y)^k (x + y) - (x^k) x - (y^k) y = x[(x + y)^k - x^k - y^k] + y[(x + y)^k - x^k - y^k] + xy^k + yx^k = x[Axy(x + y)] + y[Bxy(x + y)] + xy[x^(k - 1) + y^(k - 1)] where A and B are the expressions of variables x and y As x^(k - 1) + y^(k - 1) is divisible by (x + y), we just prove that (x + y)^(k + 1) - x^(k + 1) - y^(k + 1) is divisible by (x + y). So. P(k + 1) is true. By M.I. for all positive integer n, (x+y)^n - x^n - y^n is divisible by xy(x + y)
其他解答:
@myisland8132 sorry!!! 岩岩先睇到 "n is a positive odd integer" 即係呢個唔可以用M.I. prove.. 唔好意思!!!
mathematical induction
發問:
此文章來自奇摩知識+如有不便請留言告知
Prove by M.I. that (x+y)^n - x^n - y^n is divisible by xy(x+y). Thanks a lot. 更新: thx a lot!! can we assume "(x+y)^k - x^k - y^k = xy(x+y)N, where N is an integer" at the beginning instead of using A and B?最佳解答:
Let P(n): (x+y)^n - x^n - y^n is divisible by xy(x+y). When n = 1: (x + y) - x - y = 0 which is divisible by xy(x + y) Assume that P(k) is true, (x+y)^k - x^k - y^k is divisible by xy(x+y). When n = k + 1 (x + y)^(k + 1) - x^(k + 1) - y^(k + 1) = (x + y)^k (x + y) - (x^k) x - (y^k) y = x[(x + y)^k - x^k - y^k] + y[(x + y)^k - x^k - y^k] + xy^k + yx^k = x[Axy(x + y)] + y[Bxy(x + y)] + xy[x^(k - 1) + y^(k - 1)] where A and B are the expressions of variables x and y As x^(k - 1) + y^(k - 1) is divisible by (x + y), we just prove that (x + y)^(k + 1) - x^(k + 1) - y^(k + 1) is divisible by (x + y). So. P(k + 1) is true. By M.I. for all positive integer n, (x+y)^n - x^n - y^n is divisible by xy(x + y)
其他解答:
@myisland8132 sorry!!! 岩岩先睇到 "n is a positive odd integer" 即係呢個唔可以用M.I. prove.. 唔好意思!!!
文章標籤
全站熱搜
留言列表
