*x2n-1+y2n-1(n∈N*)能被x+y整除.
来源:语文精选馆 3.14W
问题详情:
*x2n-1+y2n-1(n∈N*)能被x+y整除.
【回答】
* (1)当n=1时,x2n-1+y2n-1=x+y,能被x+y整除.
(2)假设当n=k(k∈N*)时,命题成立,
即x2k-1+y2k-1能被x+y整除.
那么当n=k+1时,x2(k+1)-1+y2(k+1)-1
=x2k+1+y2k+1=x2k-1+2+y2k-1+2
=x2·x2k-1+y2·y2k-1+x2·y2k-1-x2·y2k-1
=x2(x2k-1+y2k-1)+y2k-1(y2-x2).
∵x2k-1+y2k-1能被x+y整除,
y2-x2=(y+x)(y-x)也能被x+y整除,
∴当n=k+1时,x2(k+1)-1+y2(k+1)-1能被x+y整除.
由(1),(2)可知原命题成立.
知识点:推理与*
题型:解答题