a-maths
回答 (2)
2007-09-06 8:03 am
參考: My Maths knowledge
2007-09-06 6:09 am
Let s(n) be the statement' 6^n (5n-1)+1=25m where m is a positive integer.'
When n=1 LHS=6^1 (5(1)-1)+1 = 6*4+1=25
RHS=25
LHS=RHS
SO That s(1) is true.
Assume that s(k) is also true.
i.e. 6^k (5k-1)+1=25 w where it is a positive integer.
When n=k+1
LHS= 6^(k+1) (5(k+1)-1)+1
= 6^k* 6^1*(5k+5-1)+1
= 6^k*6*1*(5k-1+5)+1
=1(6^k* 5k-1)+5(6*k)+5
=25w+5(6*k)+5
=5(5w+6*k+1)
As (5w+6*k+1) is a positive integer.
SO that s(k+1) is true.
When n=1 LHS=6^1 (5(1)-1)+1 = 6*4+1=25
RHS=25
LHS=RHS
SO That s(1) is true.
Assume that s(k) is also true.
i.e. 6^k (5k-1)+1=25 w where it is a positive integer.
When n=k+1
LHS= 6^(k+1) (5(k+1)-1)+1
= 6^k* 6^1*(5k+5-1)+1
= 6^k*6*1*(5k-1+5)+1
=1(6^k* 5k-1)+5(6*k)+5
=25w+5(6*k)+5
=5(5w+6*k+1)
As (5w+6*k+1) is a positive integer.
SO that s(k+1) is true.
參考: Me
收錄日期: 2021-04-23 20:32:09
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070905000051KK04389