a1=1,2an+1=(1+1/n)^2*an 证明{an/n^2}为等比数列 求{an}通项公式 令bn=（an+1）,求数列{bn}的前n项和Sn

来源：学生作业帮助网 编辑：作业帮 时间：2024/11/06 01:50:17
a1=1,2an+1=(1+1/n)^2*an 证明{an/n^2}为等比数列 求{an}通项公式 令bn=（an+1）,求数列{bn}的前n项和Sn
xTn0z&HL/d$@vfm@Z@ h@p?V)s(;As/fZ(1΄E&N*:]n.?ԚJ\4O_/Q ׿w?6h}mWH~2󛫷Mz4yş7GϛcP1E*xޕ,dQRH4hшá,P A 'c& J[e|FY~)ّvj[t@jr@GGV21W*! Ӡ!MgqdtʟYEnwܢ@c7rFf鍅:.v>C%-,NPk.t tÐ ǜZpepw<wL& <v

a1=1,2an+1=(1+1/n)^2*an 证明{an/n^2}为等比数列 求{an}通项公式 令bn=（an+1）,求数列{bn}的前n项和Sn
a1=1,2an+1=(1+1/n)^2*an 证明{an/n^2}为等比数列 求{an}通项公式 令bn=（an+1）,求数列{bn}的前n项和Sn

a1=1,2an+1=(1+1/n)^2*an 证明{an/n^2}为等比数列 求{an}通项公式 令bn=（an+1）,求数列{bn}的前n项和Sn
2a(n+1)=(1+1/n)^2a(n)
2^(n+1)a(n+1)=(n+1)^2[2^na(n)]/n^2
2^(n+1)a(n+1)/(n+1)^2=2^na(n)/n^2=...=2a(1)/1=2
a(n)=n^2/2^(n-1)
b(n)=a(n+1)=(n+1)^2/2^n=n(n+1)/2^n + (n+1)/2^n=c(n)+d(n)
c(n)=n(n+1)/2^n,d(n)=(n+1)/2^n
D(n)=d(1)+d(2)+...+d(n)=2/2 + 3/2^2 + 4/2^3 + ...+ (n-1+1)/2^(n-1) + (n+1)/2^n
2D(n)=2 + 3/2 + 4/2^2 +...+ (n-1+1)/2^(n-2) + (n+1)/2^(n-1)
D(n)=2D(n)-D(n)=2+1/2+1/2^2+...+1/2^(n-1) - (n+1)/2^n
=1-(n+1)/2^n + 2[1-1/2^n]
=3-(n+3)/2^n
C(n)=c(1)+c(2)+...+c(n)=1*2/2 + 2*3/2^2 + 3*4/2^3 + ...+(n-1)n/2^(n-1) + n(n+1)/2^n
2C(n)=1*2 + 2*3/2 + 3*4/2^2 + ...+ (n-1)n/2^(n-2) + n(n+1)/2^(n-1)
C(n)=2C(n)-C(n)=1*2 + 2(3-1)/2 + 3(4-2)/2^2 + ...+ n(n+1-n+1)/2^(n-1) - n(n+1)/2^n
=2+ 2[(1+1)/2 + (2+1)/2^2 + ...+ (n-1+1)/2^(n-1)] - n(n+1)/2^n
=2+2[D(n)-(n+1)/2^n] - n(n+1)/2^n
S(n)=C(n)+D(n)=2+2[D(n)-(n+1)/2^n] - n(n+1)/2^n + D(n)
=2-n(n+1)/2^n - 2(n+1)/2^n + 3D(n)
=2-n(n+1)/2^n -2(n+1)/2^n + 3[3-(n+3)/2^n]
=11-(n^2+6n+11)/2^n

2a(n+1)=(1+1/n)^2*an
2a(n+1)=[(n+1)/n]^2*an
2a(n+1)/(n+1)^2=an/n^2
[a(n+1)/(n+1)^2]/(an/n^2)=1/2
所以{an/n^2}是以1/2为公比的等比数列
an/n^2=(a1/1^2)*q^(n-1)
an/n^2=(1/2)^(n-1)
an=n^2*(1/2)^(n-1)

an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么意思? an=a1+(n-1)d 和Sn=n(a1+an)/2中an是什么 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 数列{an}满足a1=1 an+1=2n+1an/an+2n 数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. 数列{an}中,a1=1,an+1/an=n/n+2,求an a1=1,Sn=n^2an,求an 设a1,a2,...,an都是正数,证明不等式(a1+a2+...+an)[1/(a1)+1/(a2)+...+1/(an)]>=n^2 已知数列{An}满足A1=0.5,A1+A2+…+An=n^2An(n∈N*),试用数学归纳法证明:An=1/n(n+1) 已知A1=1,An=2An-1+n（n>1）,求An. an=an-1+2的n次方+n,a1=1,an=? 数列an满足a1=1/3,Sn=n(2n-1)an,求an 数列{an}中,a1=2,an+1=an+ln(n+1/n),求通项公式 数列{an},a1=3,an*a(n+1)=(1/2)^n,求an 数列{an}满足a1=2,a(n+1)=2an+n+2,求an a1=5,n大于等于2时,an-an-1=6n,求an 数列an,a1=1.an+1/an=n+2/n,求通项公式, 已知数列｛an｝中,a(n+1)=an+2^n,a1=3,求an