已知数列{an}的前n项和Sn=-an-(1/2)的n-1次方再加2,令bn=2的n次方乘an,求证bn等差,
还要求an
人气:218 ℃ 时间:2019-10-18 02:47:18
解答
Sn = -an + 2 - 2^(1-n)
S(n+1) = -a(n+1) + 2 - 2^(-n)
a(n+1)= -a(n+1) + an - 2^(-n) + 2^(1-n)
2a(n+1) = an + 2^(-n)
两边同乘以2的n次方
得到2^(n+1)·a(n+1) - 2^n·an = b(n+1) - bn = 1
S1 = a1 = -a1 +1 得 a1 = 0.5
b1 = 2a1 =1
bn = n = 2^n·an 得 an = 2^(-n)an=n/(2^n)呃 是我疏忽了不好意思
推荐
- 数列an中,a1=1,an+1=2an+2的n次方,设bn=an/2∧n-1,证明bn是等差数列,求数列an的前n项和sn
- 设数列an中的前n项的和为Sn,并且a1=1,Sn+1=4an+2,设bn=an比2的n次方,求证数列bn为等差数列
- 设数列(An)的前N项和为Sn,已知Sn=2An-2的n次方.(1)设(Bn)=an/2的n次方-1,证明(Bn)为等差,
- 已知等差数列前三项和为6前八项和为-4.设数列bn等于{4-an)3的n-1次方求sn
- 数列前n项和为Sn,a1=1,S(n +1)=4(an)+2.设bn=an/(2的n次方),证:(bn)是等差数列.
- ln(1/e)等于多少
- october 31st前用on还是in
- 作比较、举例子和分类别的句子
猜你喜欢