这个第三题怎么做
3个回答
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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
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
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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
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
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这题x,y不能为0,不然范围就是n大于等于22
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