an=n²(cos²nπ 3-sin²nπ 3)
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∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/2)+
an=2^(n)-1-(2^(n-1)-1)=2*(2^(n-1))-1-2^(n-1)+1=2^(n-1)你上面少个-1
(1)∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/
是cos(2nπ/3)的意思吧? 晕死.懒得重新编辑了. 按照那个思路每一项乘以cosA-sinA.应该有更简单的方法. PS.连加的意思.就是以
an=n(cosnπ/3-sinnπ/3)=n^2*cos(2nπ/3)(二倍角公式)cos(2π/3)=-1/2cos(4π/3)=-1/2cos(6π/3)=1所以a(3k-2)+a(3k-1)+
因为{an}是等差数列,所以2a7=a6+a8,2b7=b6+b8即S13=13a7,S`13=13b7所以a7/b7=S13/S`13=(7*13+2)/(13+3)=93/16
(1)根据a(1)=1、a(2)=2及a(n+2)=[1-(1/3)cos²(nπ/2)]a(n)+2sin²(nπ/2)可得a(3)=[1-(1/3)cos²(π/2)
评析:本页那位热心网友写错了:在得出an+1=3(a(n-1)+1)后,应将a2=8带入求值,因为前面a(n-1),n应大于等于二,所以a1不能算入通项公式中,应检验是否符合n大于等于二时的通项公式,
a(n)=ncos(nπ/2+π/3),n=1,2,...a(4k)=4kcos(4kπ/2+π/3)=4kcos(π/3)=2k,a(4k+1)=(4k+1)cos[(4k+1)π/2+π/3]=(
s1=a1=2a1+1^2-3*1-2a1=2a1-4a1=4sn=2an+n^2-3n-2s(n-1)=2a(n-1)+(n-1)^2-3(n-1)-2sn-s(n-1)=2an+n^2-3n-2-
利用三角函数的积化和差公式,得到an=sin(n+1)cos(n-1)/n^p=[sin(2n)+sin2]/2n^p={sin(2n)/n^p+sin2/n^p}/2可证当0再问:确实是条件收敛,我
a(n+2)=[1+cos^2(nπ/2)]an+sin^2(nπ/2)]若n为偶数a(n+2)=2an若n为奇数a(n+2)=an+1∴a1,a3,a5……形成等差数列a2,a4.26……形成等比数
(cos(nπ/3))^2-(sin(nπ/3))^2=cos(2nπ/3)n=1,cos(2π/3)=-1/2n=2,cos(4π/3)=-1/2n=3,cos(6π/3)=1以后cos取值三个一组
Sn+S(n+1)=5(a(n+1))/3因为S(n+1)=SN+A(N+1)所以Sn+SN+A(N+1)=5a(n+1)/32SN=2a(n+1)/3SN=a(n+1)/3S(N-1)=AN/3SN
(1)、an+2=[1+cos(nπ/2)]an+sin(nπ/2),n∈N因为cosnπ=2cos(nπ/2)-1=1-sin(nπ/2),所以an+2=[1+cos(nπ/2)]an+sin(nπ
(1)an=n^2cos2πn/3cos2πn/3取到的值为-1/2,-1/2,1,-1/2,-1/2,1,.对于n=3k(k∈N*),a(3k-2)+a(3k-1)+a(3k)=-1/2(3k-2)
因为a(n+1)=S(n+1)-S(n)=S(n)+3n+1即a(n+1)=S(n)+3n+1(1)所以a(n)=S(n-1)+3(n-1)+1(2)(1)-(2)得a(n+1)-a(n)=S(n)-
∵an=n[cos(nπ)/3-sin(nπ)/3]=n×cos(2π/3)n∵n取1到n时,cos(2π/3)n的取值依次为-1/2,-1/2,1,-1/2,-1/2,1,……∴S10=-1/2×(
∵数列{a[n]}的通项a[n]=n^2[(cosnπ/3)^2-(sinnπ/3)^2],前n项和为S[n]∴a[n]=n^2Cos(2nπ/3)∴S[n]=1^2(-1/2)+2^2(-1/2)+
之前你出过这种题了吧,原来让求的是前30项.也不说清楚是从a0还是a1开始,不过不要紧a0=0;之前求的是S29,S30如下cos(nπ/3)^2-sin(nπ/3)^2=1-2sin(nπ/3)^2