等比数列前n项和Sn=a3n-1次方 b
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Sn=na1,q=1a1(1-q^n)/(1-q)=(a1-anq)/(1-q),q≠1
由Sn=3^(n+1)+r可知公比q=3取n=1得a1=9+r取n=2得a1+a2=4a1=27+r解得a1=6,r=-3
a1=S1=2+ka2=S2-S1=(4+k)-(2+k)=2a3=S3-S2=(8+k)-(4+k)=4等比则a2²=a1a34=4(2+k)k=-1
设等比数列{a[n]}的公比为q则S[n]=a[1](1-qⁿ)/(1-q)=2(1-qⁿ)/(1-q)则S[n]+1=2(1-qⁿ)/(1-q)+1S[1]+1=
n2*cos(2nπ/3)是什么意思?再问:n^2乘cos(2nπ/3),再答:令Dn=A3n-2+A3n-1+A3n,在由an的通项解得Dn=(3n)^2-1/2(3n-2)^2-1/2(3n-1)
an+sn=-2n-1,当n=1时,a1+s1=-3,则a1=-3/2.由已知得:sn=-2n-1-an当n大于或等于2时,则an=sn-s(n-1)=-2n-1-an-[-2(n-1)-1-a(n-
n=1时,a1=1+3a1.即a1=-1/2.n>1时,an=Sn-Sn-1=1+3an-(1+3a(n-1))=3an-3a(n-1),即an=3/2a(n-1),即an=-1/2*(3/2)^(n
Sn=n-5an-85S1=1-5a1-85即a1=1-5a1-85解得a1=-14an=Sn-S(n-1)=n-5an-85-[(n-1)-5a(n-1)-85]=-5an+5a(n-1)+16an
设首项为a1,公比为r,当r=1时,Sn=n(a1),此时Sn/S(n+1)的极限为1r≠1时,Sn=a1(1-r^n)/(1-r),Sn/S(n+1)=(1-r^n)/(1-r^(n+1)),极限为
∵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
s(n+1)-sn=5/6(n+1)(n+4)-5/6n(n+3)=5/6(n²+5n+4-n²-4n-3)=5/6(n+1)=5/6n+5/6所以an是等差数列
a1=S1=3k+1.a2=S2-S1=9k+1-(3k+1)=6k.a3=S3-S2=(27k+1)-(9k+1)=18k.等比数列.a3/a2=3.则q=3.所以3a1=a2.即3*(3k+1)=
显然k=-1,等比数列前n项和中指数式的系数和常数项互为相反数
an=Sn-S(n-1)=k*2*3^(n-1)a1=2k,q=3,Sn=a1*(1-3^n)/(1-3)=k(3^n-1)=k*3^n+1所以k=-1
1.已知的是等比数列求{A3n},即A1,A3,A6,A9.A3n总共n项,该等比数列为Tn原数列An的公比为An/A(n-1)=1/2则可得到Tn的公比为An/A(n-3)=1/4得Tn=A1(1-
这个直接用a5=s5-s4=(32+r)-(16+r)=16
a1=S1=3+a,a2=s2-s1=9+a-(3+a)=6,a3=S3-S2=27+a-(9+a)=18,因为a1·a3=a2²,即18(3+a)=36,解得a=-1再问:求通项公式再答:
已知Sn=2An-1取n=1得:S1=2A1-1又因为S1=A1,解上述方程可得:A1=1Sn=2An-1S(n-1)=2A(n-1)-1注:"n-1"为下标上下两式相减得:Sn-S(n-1)=2An
(1)令n=1,得a1=-1.Sn=2an+n,S(n+1)=2a(n+1)+n+1.两式相减,得a(n+1)=2a(n+1)-2an+1.整理得a(n+1)-1=2(an-1),a1-1=-2.综上
a1=s1=5-aan=sn-s(n-1)=5^n-a-(5^(n-1)-a)=5^n-5^(n-1)=4*5^(n-1)当n=1时,an=4即5-a=4a=1