题目内容
(1)若J1输入1,J2输入正整数n,输出结果为多少?
(2)若J2输入1,J1输入正整数m,输出结果为多少?
(3)若J1输入正整数m,J2输入正整数n,输出结果为多少?
分析:(1)由题意,可得f(1,n)}成等差数列,公差为2,首项为f(1,1)=1,从而可得f(1,n);
(2){f(m,1)}为等比数列,公比为2,首项为f(1,1)=1,从而可得f(m,1);
(3)f(m,1)看作是数列的首项,f(m,n+1)=f(m,n)+2,这里n+1,n相当于数列的项数,2相当于数列的公差.从而可得f(m,n).
(2){f(m,1)}为等比数列,公比为2,首项为f(1,1)=1,从而可得f(m,1);
(3)f(m,1)看作是数列的首项,f(m,n+1)=f(m,n)+2,这里n+1,n相当于数列的项数,2相当于数列的公差.从而可得f(m,n).
解答:解:(1)由题意得f(1,1)=1,f(m,n+1)-f(m,n)=2,
=2∴{f(1,n)}成等差数列,公差为2,首项为f(1,1)=1∴f(1,n)=f(1,1)+(n-1)2=2n-1(14分)
(2){f(m,1)}为等比数列,公比为2,首项为f(1,1)=1
∴f(m,1)=f(1,1)2m-1=2m-1(4分)
(3)∵f(m,1),f(m,2),…,f(m,n)成等差数列,公差为2,首项f(m,1)=2m-1
∴f(m,n)=f(m,1)+2(n-1)=2m-1+2(n-1)(6分)
| f(m+1,1) |
| f(m,1) |
(2){f(m,1)}为等比数列,公比为2,首项为f(1,1)=1
∴f(m,1)=f(1,1)2m-1=2m-1(4分)
(3)∵f(m,1),f(m,2),…,f(m,n)成等差数列,公差为2,首项f(m,1)=2m-1
∴f(m,n)=f(m,1)+2(n-1)=2m-1+2(n-1)(6分)
点评:本题解题的思想是类比特征,看作是数列问题,利用数列知识求解.
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