an=|3n-12|求Sn

sn=2*3^n,求anSn=2*3^n=3*2*3^(n-1)Sn-1=2*3^(n-1)an=Sn-Sn-1=2*2*3^(n-1)=4*3^(n-1)an=4*3^(n-1)

已知an=(2n+1)*3^n,求Snan=(2n+1)*3^na1=3*3^1a2=5*3^2a3=7*3^3.an=(2n+1)*3^nSn=3*3^1+5*3^2+7*3^3+.(2n+1)*3^n3Sn=3*3^2+5*3^3+7*

数列求和an=n（n+3）求SnSn=a1+a2+a3+……+an=1（1+3）+2（2+3）+3（3+3）+……+n（n+3）=1^2+2^2+3^2+……n^2+3(1+2+3+……+n)=n（n+1)（2n+1）/6+3n（n+1）/

Sn=n平方-n.求anan=sn-sn-1=n^2-n-(n-1)^2+(n-1)=n^2-n-n^2+2n-1+n-1=2n-2an=Sn-Sn-1=n^2-n-[(n-1)^2-(n-1)]=2n-2An=(Sn)-S(n-1)=n&

an=n*2^n,求SnSn=1×2＋2×2²＋3×2³＋…＋n×2^n2Sn=1×2²＋2×2³＋3×2^4＋…＋(n－1)×2^n＋n×2^(n＋1)两式相减,得：－Sn=2＋2²＋2&

an=n(n+1)求snan=n^2+nn^2对应的前n项和为n(n+1)(2n+1)/6(这个作为一个结论记住就可以了属于性质类的)n对应的前n项和为n(n+1)/2sn=n(n+1)(2n+1)/6+n(n+1)/2

an=2^n+n,求Snan=2^n+nSn=2+2^2+2^3+……+2^n+1+2+3+……+n=2(1-2^n)/(1-2)+(1+n)n/2=2(2^n-1)+(1+n^2)/2

an=n+2^（n-1）,求snan=3/n（n+1）,求snan=1an=n+2^（n-1）,求snan=3/n（n+1）,求snan=1/（2n+1）（2n-1）求sn

数列{an}前n项和为Sn,且2Sn+1=3an,求an及Sn当n=1时、有2s1+1=3a1,即有a1=1,因为2Sn+1=3an,所以2Sn+1+1=3an+1.后式减去前式,得2an+1=3an+1-3an.即有an+1=3an,为等

已知数列{an}的的前n项和Sn=3n²-12n      (1)求an        (

数列an满足a1=1/3,Sn=n(2n-1)an,求anSn=n(2n-1)an,S(n+1)=(n+1)[2(n+1)-1]a(n+1)=(n+1)[2n+1]a(n+1)S(n+1)-Sn=a(n+1)即(n+1)[2n+1]a(n+

a1=1,an+1=an+3^n-n,求an与snan=a(n-1)+3^(n-1)-(n-1)=...=3^(n-1)+3^(n-2)+...+3^1-(n-1)-(n-2)-...-1+a1=(3-3^n)/(1-3)-(n-1)*n/

数列｛an｝中①Sn=π（2n²+n）/12②Sn=2/3（3n-1）求an(1)Sn=π(2n^2+n)/12(1)S(n-1)=π(2(n-1)^2+(n-1))/12(2)(1)-(2)an=(π/12)(4n-1)(2)S

数列an,an=(2n-1)+1/【3n(n+1)】,求Snan=(2n-1)+1/【3n(n+1)】令bn=2n-1cn=1/【3n(n+1)】=1/3*[1/n-1/（n+1)]则sn=sbn+scn很简单。分组求和法。

等差数列{an}前n项和为Sn=3n-2n^2,求anan=sn-s(n-1)这个公式挺常用的,用这个直接就解出来了所以an=3n-2n^2-[3(n-1)-2(n-1)^2]右边化简,得an=3n-2n^2-[3n-3-2(n^2-2n+

Sn为等比数列{an}前n项和,an=(2n-1)*3的n次方,求SnSn=3^1+3×3^2+5×3^3+……+(2n-1)×3^n①3Sn=3^2+3×3^3+5×3^3+……+(2n-1)×3^n+1②①-②得：-2Sn=3^1+2×

已知数列AN的前N项和SN,SN=2N^2+3n+2,求anA(n+1)=S(n+1)-Sn=2(n+1)^2+3(n+1)+2-2n^2-3n-2=2n^2+4n+2+3n+3-2n^2-3n=4n+5An=5+4(n-1)当n=1时，a

An=1/n,求Sn.1+1/2+1/3+……+1/n是调和级数,没有确定的表达式!但它存在极限值,当n趋于无穷大时,其极限值为ln2,.

数列｛an｝,中,a1=1/3,设Sn为数列｛an｝的前n项和,Sn=n（2n-1）an求Sn得s1为1／3,则s1/a1为1,则可知sn/an＝n（2n-1）,有因为an＝sn-sn-1,所以得sn/sn-sn-1＝2n-n,化简得sn／

已知数列前n项和为Sn,且Sn=-2n+3,求an及Snan=sn-s(n-1)=-2n+3-[-2(n-1)+3]=-2sn为已知.强烈怀疑你抄错题了给点分呀n=1s1=1即a1=1;n=2s2=-1即a1+a2=-1,a2=-2n=3s