Sn=2n-n
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有条件a1=2,d=2吧,an=2n,S1=a1=1*(1+1),其满足,假设Sj=j^2+j=j(j+1),而a(j+1)=2(j+1),则S(j+1)=Sj+a(j+1)=(j+1)(j+2),满
分析:由于对于数列的n值有不同范围取值,对应不同的求和公式,可知数列为分段数列,需要对不同范围的n值进行讨论,方可求得数列的通项公式;当n=1时,a1=S1=3+1=4;当2≤n≤5时,an=Sn-S
Sn+1/(2n+1)-Sn/(2n-1)=1Sn/(2n-1)=S1+n-1→Sn=(S1+n-1)(2n-1)→Sn=n(2n-1)an=4n-31/√an=2/2√(4n-3)>2/(√4n-3
用错位相减法a1=1*2^0a2=2*2^1a3=3*2^2.an=n*2^(n-1)Sn=1*2^0+2*2^1+3*2^2+.+n*2^(n-1)2Sn=1*2^1+2*2^2+3*2^3+.+(
由题得:an=1/2(1/n-1/(n+1);所以:a1=1/2(1-1/2);a2=1/2(1/2-1/3);a3=1/2(1/3-1/4);.an=1/2(1/n-1/(n+1);sn=a1+a2
我来试试吧.Sn+1=1*(n+1)+2*(n)+3*(n-1)+……+(n+1)*1=1*n+1+2*(n-1)+2+3*(n-2)+3+……+n*1+n=1*n+2*(n-1)+3*(n-2)+…
我就说第二问吧.若{an}中存在三项,它们可以构成等差数列,则有2an=(an-1)+(an+1)即2*(3*2^n-3)=3*2^(n+1)-3+3*2^(n-1)-3,3*2^(n+1)-6=3*
an=(2^n-1)n=2^n*n-n,令Tn=2^1*1+2^*2+…2^n*n,①则2Tn=2^2*1+2^3*2+…+2^n*(n-1)+2^(n+1)*n②②-①得Tn=-(2^1+2^2+…
∵a(n+1)=(n+2)Sn/n且a(n+1)=S(n+1)-Sn∴S(n+1)-Sn=(n+2)*Sn/n∴S(n+1)=[(n+2)/n+1]Sn=(2n+2)/n*Sn∴S(n+1)/(n+1
an=(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
(1)证明:由Sn=2an-3n,得Sn-1=2an-1-3(n-1)(n≥2),则有an=2an-2an-1-3an+3=2(an-1+3)(n≥2),∵a1=S1=2a1-3,∴a1=3,∴a1+
Sn=(3n+1)/2-(n/2)an当n=1时,a1=4/3=1+1/3=1+1/[1*(1+2)]当n=2时,a2=13/12=1+1/[2*(1+2+3)当n=3时,a3=31/30=1+1/[
一,n为奇数,Sn=nC(n,n)+(1+n-1)C(n,1)+(2+n-2)C(n,2)+…+nC(n,n-1/2)=n[C(n,0)+C(n,1)+…+C(n,n-1/2)=n*2de(n-1)次
等于呀,你把后面的算式一道前面来n(n+2)(n+4)+1/6)(n-1)n(n+2)(n+4)=n(n+2)(n+4)[1+1/6(n-1)]=1/6n(n+2)(n+4)(n+5)
再问: 再问:那个划横线的答案是不是错了再答:我觉得是
Sn=2+5n+8n^2+…+(3n-1)n^n-1nSn=2n+5n^2+…+(3n-4)n^(n-1)+(3n-1)n^nSn-nSn=2+3n+3n^2+…+3n^(n-1)-(3n-1)n^n
sn=a1+a2+a3+.+an=(1^2+2^2+3^2+.+n^2)-(1+2+3+...+n)+2n=n(n+1)(n+2)/6-n(1+n)/2+2n再问:三次方?这是什么数列?再答:an=n
(1)An=3(1+2^n)(2)由题知,Sn=2An+3n-12=6(2^n-1)+3nBn=(An-3)/(Sn-3n)(A(n+1)-6)=(3*2^n)/(6(2^n-1))(3(2^(n+1
裂项相消:Cn=[(n+2)/n-(n+2)/(n+1)]2^n=2/(n2^n)-1/((n+1)2^n)=1/(n2^(n-1)-1/((n+1)2^n),因此Sn=1-1/(2*2)+1/(2*
an=sn-sn-1=n^2+3n-(n-1)^2-3(n-1)=2n-1+3=2(n+1)an-an-1=2(n+1)-2n=2所以为等差数列