|
问题：【{an}为等差数列求证（1）aka（2k）a（3k）构成等差数列（2）a1+an=a（1+k）=a（n-k）（3）SkS（2k）-SkS（3k）-S（2k）构成等差数列】
答案：↓↓↓ 宋李俊的回答： 网友采纳 {an}为等差数列 假设an=a1+(n-1)d,d为公差,a1为第一项 则ak=a1+(k-1)d a(2k)=a1+(2k-1)d a(3k)=a1+(3k-1)d a(2k)-a(k)=[a1+(2k-1)d]-[a1+(k-1)d]=kd a(3k)-a(2k)=[a1+(3k-1)d]-[a1+(2k-1)d]=kd 所以得证等差 （2）题目写错了,应该是a1+an=a（1+k）+a（n-k） a（1+k）+a（n-k）=a1+kd+a1+(n-k-1)d =2a1+(n-k-1+k)d =2a1+(n-1)d a1+an=a1+a1+(n-1)d =2a1+(n-1)d 所以得证 （3）Sk=(a1+ak)k/2=(2a1+(k-1)d)k/2 S(2k)=(2a1+(2k-1)d)2k/2 S(3k)=(2a1+(3k-1)d)3k/2 所以[S(2k)-Sk]-Sk=[(2a1+(2k-1)d)2k/2-(2a1+(k-1)d)k/2]-（2a1+(k-1)d)k/2 =[(2a1+(2k-1)d)2k/2-(2a1+(k-1)d)2k/2] =（2k-1)dk-(k-1)dk =k^2*d [S(3k)-S(2k)]-[S(2k)-Sk]=[(2a1+(3k-1)d)3k/2-(2a1+(2k-1)d)2k/2]-[(2a1+(2k-1)d)2k/2-(2a1+(k-1)d)k/2] =k^2*d 所以等差 | 
