{an}为等差数列 求证(1)ak a(2k) a(3k) 构成等差数列 (2)a1+an=a(1+k)=a(n-k)(
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{an}为等差数列 求证(1)ak a(2k) a(3k) 构成等差数列 (2)a1+an=a(1+k)=a(n-k)(3)Sk S(2k)-Sk S(3k)-S(2k) 构成等差数列
{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
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
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