{an}為等差數(shù)列
假設(shè)an=a1+(n-1)d,d為公差,a1為第一項(xiàng)
則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
所以得證等差
（2） 題目寫錯(cuò)了,應(yīng)該是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
所以得證
（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