已知数列{An}、{Bn}满足a1=1/2 b1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm
来源:学生作业帮 编辑:搜狗做题网作业帮 分类:数学作业 时间:2024/05/11 13:40:32
已知数列{An}、{Bn}满足a1=1/2 b1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm+Bn
已知数列{An}、{Bn}满足A1=1/2 B1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm+Bn
(1)求数列{An}{Bn}的通项公式
(2)求数列{AnBn}的前n项和Tn
(3)若数列{Cn}满足Bn=(4Cn+n)/(3Cn+n),试求{Cn}的通项公式并判断:是否存在正整数M,使得对任意n∈N+,Cn
已知数列{An}、{Bn}满足A1=1/2 B1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm+Bn
(1)求数列{An}{Bn}的通项公式
(2)求数列{AnBn}的前n项和Tn
(3)若数列{Cn}满足Bn=(4Cn+n)/(3Cn+n),试求{Cn}的通项公式并判断:是否存在正整数M,使得对任意n∈N+,Cn
(1)A2=A1*A1=1/4,A2/A1=1/2;
An+1=An*A1,An+1/An=A1=1/2;所以An为等比数列,An=1/(2^n).
B2=B1+B1=-1,B2-B1=-1/2;
Bn+1=Bn+B1,Bn+1-Bn=B1=-1/2;所以Bn为等差数列,
Bn=-1/2+(n-1)(-1/2)=-n/2.
(2)记Dn=AnBn,Tn=D1+D2+D3+...+Dn
=-(1/2)*(1/2)-(2/2)*(1/4)-(3/2)*(1/8)-(4/2)*(1/16)-...-(n/2)*(1/2^n);
2Tn==-(1/2)*1-(2/2)*(1/2)-(3/2)*(1/4)-(4/2)*(1/8)-...-(n/2)*(1/2^n-1);
2Tn-Tn=-[1/2+1/4+1/8+1/16+...+1/2^n]+(n/2)*(1/2^n)
=-[1-1/2^n]+(n/2)*(1/2^n)=-1+(n/2+1)*(1/2^n);
即:Tn=-1+(n/2+1)*(1/2^n);(此方法称为错位相减法)
(3)-n/2=(4Cn+n)/(3Cn+n),-3nCn-n^2=8Cn+2n,Cn=-(n^2+2n)/(3n+8);
令3n+8=t,因为n>=1,t>=11;所以:
Cn=-[(1/9)(t^2-16t+64)+2(t-8)/3]/t
=-(t+16/t-10)/9;由于t+16/t当t>=11时是递增的,所以Cn就是递减的,所以t=11时Cn最大,即n=1时Cn最大,所以:Cn
An+1=An*A1,An+1/An=A1=1/2;所以An为等比数列,An=1/(2^n).
B2=B1+B1=-1,B2-B1=-1/2;
Bn+1=Bn+B1,Bn+1-Bn=B1=-1/2;所以Bn为等差数列,
Bn=-1/2+(n-1)(-1/2)=-n/2.
(2)记Dn=AnBn,Tn=D1+D2+D3+...+Dn
=-(1/2)*(1/2)-(2/2)*(1/4)-(3/2)*(1/8)-(4/2)*(1/16)-...-(n/2)*(1/2^n);
2Tn==-(1/2)*1-(2/2)*(1/2)-(3/2)*(1/4)-(4/2)*(1/8)-...-(n/2)*(1/2^n-1);
2Tn-Tn=-[1/2+1/4+1/8+1/16+...+1/2^n]+(n/2)*(1/2^n)
=-[1-1/2^n]+(n/2)*(1/2^n)=-1+(n/2+1)*(1/2^n);
即:Tn=-1+(n/2+1)*(1/2^n);(此方法称为错位相减法)
(3)-n/2=(4Cn+n)/(3Cn+n),-3nCn-n^2=8Cn+2n,Cn=-(n^2+2n)/(3n+8);
令3n+8=t,因为n>=1,t>=11;所以:
Cn=-[(1/9)(t^2-16t+64)+2(t-8)/3]/t
=-(t+16/t-10)/9;由于t+16/t当t>=11时是递增的,所以Cn就是递减的,所以t=11时Cn最大,即n=1时Cn最大,所以:Cn
已知数列{An}、{Bn}满足a1=1/2 b1=-1/2 且对任意m、n∈N+,有Am+n=Am·An,Bm+n=Bm
已知数列{an}满足a1=0,a2=2,且对任意m、n∈N*都有a2m-1+a2n-1=2am+n-1+2(m-n)2
已知数列{an}满足a1=1 ,a3=3,且对任意m,n∈N﹢都有am-1+a2n-1=2am+n-1求a2,a4.
已知等比数列{an}的通项公式为an=3^(n-1),设数列{bn}满足对任意自然数n都有b1/a1+b2/a2+b3/
已知数列{an}满足an+Sn=n,数列{bn}满足b1=a1,且bn=an-a(n-1),(n≥2),试求数列{bn}
已知数列an中,a1=1,a2=0,对任意正整数n,m(n>m)满足(an)²-(am)²=(an-
阅读下列解题过程,am+an+bm+bn=(am+an)+(bm+bn)=a(m+n)+b(m+n)=(m+n)(a+b
已知等比数列{an}的通项公式为a=3^(n-1),设数列{bn}满足对任意自然数N都有(b1/a1)+(b2/a2)+
(m+n)(a+b)=am+bm+an+bn中m,n各表示什么?
若数列an满足a1=1/3,且对任意正整数m,n都有am+n=am*an.设前n项和为sn,则s10-s9的值是?
若数列an满足a1=1/3,且对任意正整数m,n都有am+n=am*an.设前n项和为sn,则s10-s9等于?
已知数列{an}的前n项和Sn=n2(n∈N*),数列{bn}为等比数列,且满足b1=a1,2b3=b4