Bn=2n-1 2的n次方的前n项和

因为an=2n所以bn=2n×3的n次方∴Sn=2*3+2×2*3＾2+2*3*3＾3+……+2*n*3＾n两边同时除以21/2Sn=3+2*3＾2+……+n*3＾n⑴3/2Sn=3＾2+2*3＾3+

n=n×2^(n-1)Sn=b1+b2+b3+...+bn=1×2^0+2×2^1+3×2^2+...+n×2^(n-1)2Sn=1×2^1+2×2^2+...+(n-1)×2^(n-1)+n×2^n

n=2n/2^nSn=b1+b2+b3+.+b(n-1)+bn=2·1/2+2x2·1/2²+2x3·1/2³+.+2n·1/2^n.①1/2·Sn=2·1/2²+2x2

letS=1.2^1+2.2^2+...+n.2^n(1)2S=1.2^2+2.2^3+...+n.2^(n+1)(2)(2)-(1)S=n.2^(n+1)-(2^1+2^2+...+2^n)=n.2

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

分组求和Tn=b1+b2+b3+.+bn=(2*1+2^1)+(2*2+2^3)+(2*3+2^5)+.+[2n+2^(2n-1)]=(2+2*2+2*3+.+2n)+(2^1+2^3+2^5+.+2

∵Sn=n(3n-9)/2.∴bn=Sn-S(n-1)=3n-6.即k*3^n≥3n-6.化简得,k≥(3n-6)/3^n.接下来我们可以用画图的方法或者求导数的方法来做,在这里我用后者来做.令F（x

先求an令n=1,a1=s1=1；当n>=2时,an=Sn-Sn-1=(n-2)^2-(n-3)^2（注a^b表示a的b次方）=2n-5（注意,数列an不是一个等差数列,首项不符合上面的通项公式,只是

Sn=2^(n-1)-2n＝1时,a1＝S1＝-1n≥2时,an＝Sn-S(n-1)＝2^(n-1)-2^(n-2)＝2^(n-2)bn=2n+anb1＝2+a1＝1n≥2时,bn＝2n+2^(n-2

数列an的前n项和Sn=n^2-2nn=1时a1=S1=1-2=-1n>=2时an=Sn-S(n-1)=(n^2-2n)-((n-1)^2-2(n-1))=2n-3n=1时,满足an=2n-3∴an=

a1=s1=2当n>1时：Sn=n^2+nSn-1=(n-1)^2+(n-1)an=Sn-Sn-1=2n当n=1时,成立；所以an=2nbn=(1/2)^2n+n=(1/4)^n+n令Cn=(1/4)

1.n=1时,a1=S1=2n≥2时,an=Sn-S(n-1)=2ⁿ-2^(n-1)=2^(n-1)n=1时,a1=2^0=1,不满足通项公式数列{an}的通项公式为an=2n=12^(n

由于bn=(2n-1)*[(4/5)^n]则:b(n+1)=[2(n+1)-1]*[(4/5)^(n+1)]=(2n+1)*[(4/5)^(n+1)]=[(8n+4)/5]*[(4/5)^n]则:b(

由Sn=2n-n^2可得Sn-1=2(n-1)-(n-1)^2Sn-Sn-1=an=3-2nbn=5^(3-2n)=5*(1/25)^(n-1)所以｛bn｝是以5为首项1/25为公比的等比数列数列{b

n=n(n+1)=n^2+nSn=b1+b2+...+bn=(1^2+1)+(2^2+2)+...+(n^2+n)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+1)

n=2/[n*(n-1)]=2*[1/(n-1)-1/n]当n=1时,b1不可能符合bn=2/[n*(n-1)]所以n>=2时,才有bn=2/[n*(n-1)]Sn=b1+b2+b3+……+b(n-1

平方和的公式为S＝n(n+1)(2n+1)/6所以,Sn＝4×n(n+1)(2n+1)/6＋4×n(n+1)/2＝2n(n+1)(2n+1)/3＋2n(n+1)＝2n(n+1)(2n+4)/3＝4n(

An/Bn=(2n-1)/2^nSn=S1+S2+S3+……+Sn=(2*1-1)/2^1+(2*2-1)/2^2+(2*3-1)/2^3+……(2*n-1)/2^n=(1/2^0-1/2^1)+(2

错位相减法bn=1/4{n*（-2）^(n-1)}

1an=Sn-Sn-1=3^n-3^(n-1)=2*3^(n-1)2bn+1=bn+(2n-1)bn=bn-1+(2n-3)..b2=b1+1b1=-1Sbn=Sbn-1-1+[1+(2n-3)](n