{an}是等差数列且公差为d,{a2n-1+a2n}的公差?

题目

{an}是等差数列且公差为d,{a2n-1+a2n}的公差?
{a2n-1+a2n}的公差=a2n-1+a2n-(a2n-3+a2n-2)=(a2n-1-a2n-3)+(a2n-a2n-2)=2d+2d
a2n-1-a2n-3为什么等于2n?

答案

a(n)=a+(n-1)d,
a(2n-1)+a(2n)=a+(2n-1-1)d+a+(2n-1)d = 2a +(2n-2)d + (2n-2)d + d
= 2a+d + (n-1)(4d),
{a(2n-1)+a(2n)}是首项为2a+d,公差为4d的等差数列.