> 数学 >
已知{an}为等差数列,Sn为其前n项喝,且2Sn=an+2n²(1)求an.Sn
人气:159 ℃ 时间:2020-03-28 01:33:17
解答
2a(1)=2s(1)=a(1)+2,a(1)=s(1)=2,2s(n)=a(n)+2n^2,2s(n+1)=a(n+1)+2(n+1)^2,2a(n+1)=2s(n+1)-2s(n)=a(n+1)-a(n) + 2(2n+1),a(n+1) = - a(n) + 4n + 2,a(n+1) - 2(n+1) = -a(n) + 2n = -[a(n)-2n],{a(n)-2n}是首项为a...
推荐
猜你喜欢
© 2025 79432.Com All Rights Reserved.
电脑版|手机版