用第一数学归纳法证明1.(n)(n+1)(n+2)可被6整除2.(n)(n+1)(n+2)(5n+3)可被24整除只要能说明解题重点就可

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用第一数学归纳法证明1.(n)(n+1)(n+2)可被6整除2.(n)(n+1)(n+2)(5n+3)可被24整除只要能说明解题重点就可
用第一数学归纳法证明
1.(n)(n+1)(n+2)可被6整除
2.(n)(n+1)(n+2)(5n+3)可被24整除
只要能说明解题重点就可

用第一数学归纳法证明1.(n)(n+1)(n+2)可被6整除2.(n)(n+1)(n+2)(5n+3)可被24整除只要能说明解题重点就可
1、n=1时,1*2*3=6能被6整除
假设n=k时k(k+1)(k+2)能被6整除
n=k+1时,(k+1)(k+2)(k+3) = k(k+1)(k+2) +3(k+1)(k+2)
k(k+1)(k+2)能被6整除
(k+1)(k+2)能被2整除,3(k+1)(k+2)能被6整除
所以(k+1)(k+2)(k+3) 能被6整除
所以命题成立
2、n=1成立,验证略
假设n=k,k(k+1)(k+2)(5k+3)能被24整除
(k+1)(k+2)(k+3)[5(k+1)+3]
=(k+1)(k+2)(k+3)[(5k+3)+5]
=(k+1)(k+2)[(k+3)(5k+3)+5k+15)]
=(k+1)(k+2)[(5k+3)(k+4)+12]
=(k+1)(k+2)(5k+3)(k+4) +12(k+1)(k+2)
(k+1)(k+2)(5k+3)能被24整除,
(k+1)(k+2)能被2整除,12(k+1)(k+2)能被24整除
所以(k+1)(k+2)(k+3)[5(k+1)+3]
所以命题成立