知数列{an}的前n项和sn=n2(2二次方)+10n,数列bn(小n)=|an|求数列{bn}的通项公式
人气：364 ℃ 时间：2020-05-12 12:04:00
解答
因为S(n) = n^2 + 10nS(n+1) = (n+1)^2 + 10(n+1)= n^2 + 2n + 1 + 10n + 10= n^2 + 12n + 11a(n+1) = S(n+1) - S(n)= n^2 + 12n + 11 - n^2 - 10n= 2n + 11= 2(n + 1) + 9所以a(n) = 2n + 9又因为 b(n) = |a(n)|所...
推荐
- 数列｛an｝的前n项和Sn＝10n－n2,又bn＝｜an｜,求｛bn｝的前n项和.
- 设数列an的前n项和为sn,sn=n^2+n,数列bn的通项公式bn=x^(n-1)
- 已知数列an的前n项和Sn=n2+n,数列bn满足,bn=1,bn+1=2bn+1,1.求数列an和bn的通项公式
- 设数列{an}满足a1=2,an+1-an=3*2^2n-1 （1）求数列{an}的通项公式 （2）b=nan,求数列{bn}的前n项和Sn
- 设数列{An}的前n项的和为Sn已知A1=a A(n+1)=Sn+3^n (1)设Bn=Sn-3^n 求数列{Bn}的通项公式?
- 帮忙配个情侣的签名：When I am out ,take me as the wind.字数意思都相对的,
- 请帮翻译成中文 nine six nine zero eight eight nine six two
- I first poured some oil into the pot,waited for a moment and then put the vegetables into it.翻译
猜你喜欢
