{an}an>0的a1=1前n项和Sn,Sn-Sn-1=根号Sn 根号Sn 1

设：(An+1)+p(n+1)+q=4[An+pn+q]解得p=-1,q=0即An+1=4An-3n+1等价于(An+1)-（n+1）=4（An-n）若设Bn=An-n则Bn+1=4Bn则Bn=B1*

an+an-1+2n-1=0an+n=-(an-1+(n-1))[an+n]/[an-1+(n-1)]=-1an+n是等比数列,首项a+1=4,q=-1an+n=4*(-1)^(n-1)an=4*(-

因为An=Sn-Sn-1.所以Sn-Sn-1+Sn*Sn-1=0,等式两边同时除以Sn*Sn-1得：1/Sn-1/Sn-1+=1,所以1/Sn为等差数列.因为a1=1/2.所以S1=1/2,1/S1=

an-ana(n+1)-a(n+1)=0an-a(n+1)=ana(n+1)等式两边同除以ana(n+1)1/a(n+1)-1/an=1,为定值.1/a1=1/1=1数列{1/an}是以1为首项,1为

S1=a1=1/3*a1-2a1=-3a(n)=S(n)-S(n-1)=(1/3)*[a(n)-a(n-1)](2/3)*a(n)=-1/3*a(n-1)a(n)=-1/2*a(n-1)等比数列,公比

证：a(n+1)=2an/(an+1)1/a(n+1)=(an+1)/(2an)=(1/2)(1/an)+1/21/a(n+1)-1=(1/2)(1/an)-1/2=(1/2)(1/an-1)[1/a

a1=1,an+1==(-1)的n次方乘以an所以,a2=-1,a3=-1,a4=1,a5=1,a6=-1,a7=-1,a8=1,a9=1...由此看出,除a1外,依次往后四项的和为0.所以,S200

是a(n+1)=2an/(an+1)吧a(n+1)=2an/(an+1)1/a(n+1)=(an+1)/(2an)=(1/2)(1/an)+(1/2)1/a(n+1)-1=(1/2)(1/an)-(1

当n≥时an=sn-s（n-1）于是sn=n（2n-1）[sn-s（n-1)]得（2n+1）（n-1）sn=n（2n-1）s（n-1）变形为[（2n+1）/n]sn=[（2n-1）/（n-1）]s（n

1.a1=1,a(n+1)=an+3,a2=4a3=7a4=10a5=132.a1=2,a(n+1)=2ana2=4a3=8a4=16a5=323.a1=3,a2=6,a(n+2)=a(n+1)-an

an+1=2Snan-1=2Sn-1an+1-an-1=2anan=(-1)^(n+1)Sn=1/2+1/2*(-1)^(n+1)看懂了给我满意,没有别的要求,

前N项的和Sn加上第n+1项An+1,当然是前n+1项的和Sn+1咯

（1）当n≥2时，an=Sn-Sn-1=n（2n-1）an-（n-1）（2n-3）an-1，得anan−1＝2n−32n+1∴a2a1a3a2a4a3a5a4…an−1an−2anan−1＝15×37

An+2Sn*Sn-1=0Sn-Sn-1+2Sn*Sn-1=01/Sn-1-1/Sn+2=01/Sn=2nSn=1/2n(n>=2)An=1/(2n-2n^2)(n>=2)=1/2(n=1)

2Sn=(n+1)an2S（n-1）=na（n-1）两式相减得2an=(n+1)an-na（n-1）移相得（1-n）an=-na（n-1）得an=(n/(n-1))a(n-1)an=(n/(n-1))

设公比是qan+2an+1+an+2=0∴an+2an*q+an*q²=0∴an(1+2q+q²)=0∵an≠0∴1+2q+q²=0∴(q+1)²=0∴q=-1

（1）S1=a1=（2a1/a1）-1=1S2=2a2/a1-1=2a2-1=a1+a2=1+a2所以2a2-1=1+a2a2=2（2）Sn=(2an/a1)-1=2an-1Sn-1=（2an-1/a

/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)

an+2SnSn-1=0Sn-Sn-1+2SnSn-1=01/Sn-1/Sn-1=21/Sn=2+2(n-1)Sn=1/nan=Sn-Sn-1=1/n-1/(n-1)1/2n=1an=-1/[n(n-

an－a(n＋1)=ana(n＋1)【两边同除以ana(n＋1)】得：1/[a(n＋1)]－1/[a(n)]=1即：数列{1/(an)}是以1/a1=1为首项、以d=1为公差的等差数列.则：1/[a(