Question

这个第三题怎么做

Answer 1

用归纳法
证：
n=34时，n=3x4+11x2(n可分解为3x+11y的形式），n∈A
n=35时，n=3x8+11x1，n∈A
n=36时，n=3x1+11x3，n∈A
假设n=k，n=k+1，n=k+2时，n∈A
则k=3x+11y(x，y∈N*)
k+3=3x+11y+3=3(x+1)+11y，k+3∈A
同理k+4∈A，k+5∈A
如果n=k，n=k+1，n=k+2时，n∈A，
那么n=k+3，n=k+4，n=k+5时，也有n∈A

归纳证明n≥34且n∈N*时，n∈A

Answer 2

A={ 3x+11y | x,y∈N+}
n≥34 , n∈N+
By MI
n=34
x=1, y=11∈N+
34 = 3(1)+11(3)
=> 34 ∈ A
Assume p(k) is true , k≥34
ie
∃ n, m ∈N+
k = 3n +11m
for n=k+1
k+1 = 3n +11m +1 ( By assumption )
case 1: m=1
k+1 = 3n +11m +1
= 3(n-7) +11m +22
=3(n-7) +11(m+2) ∈A
case 2: m>1
k+1 = 3n +11m +1
= 3(n-1) +11(m-1) +15
= 3(n+4) +11(m-1) ∈A
=> By principle of MI, it is true for all n≥34

Answer 3

这题x，y不能为0，不然范围就是n大于等于22