设数列an对所有正整数n都满足a1 2a2 2²a3
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由:a1+2a2+2^2a3+…+2^(n-1)an=8-5n--------------------------------①知:a1+2a2+2^2a3+…+2^(n-2)a(n-1)=8-5(n
请问是am+n中是m+n是下标还是只有m是下标?如果是m+n是下标,则可设m=1则an+1=an×a1=an/3∴后一项是前一项的1/3倍,则这是以1/3为公比,1/3为首项的等比数列.∴Sn=1/2
令m=1a(n+1)=a1*an则an是以a1为公比的等比数据列an=1/3^nS10-S9=a10=1/3^10
an=Sn-Sn-1=n(a1+an)/2-(n-1)(a1+an-1)/22an=na1+nan-na1-nan-1+a1+an-1(n-2)an=(n-1)*(an-1)-a1(1)同理(n-1)
因为an=5Sn+1所以a(n-1)=5S(n-1)+1所以an-a(n-1)=5Sn+1-[5S(n-1)+1]所以an-a(n-1)=5[Sn-S(n-1)]=5an所以an/a(n-1)=-1/
a1×a2×a3×a4=a1+a2+a3+a41×1×2×a4=1+1+2+a4a4=4a2×a3×a4×a5=a2+a3+a4+a51×2×4×a5=1+2+4+a57a5=7a5=1=a1a3×a
a1^3+a2^3+...+an^3=sn^2a1^3+a2^3+...+[a(n+1)]^3=[s(n+1)]^2两式相减得[a(n+1)]^3=[s(n+1)]^2-sn^2[a(n+1)]^3=
(1)bn,√an,bn+1成等比所以an=bn*bn+1所以a1=b1*b2=3a2=b2*b3=6所以b1*(b1+d)=3(b1+d)*(b1+2d)=6解得:b1=√2d=√2/2或者b1=-
当n=1时,有a2/a1=(4*1-1)/(2*1-1)=3,∴a2=3a{an}不是等差数列吗?那好,公差d=a2-a1=2a∴an=a1+(n-1)*d=a*(2n-1),n∈N*再问:谢谢了,还
1、①A1+3A2+3^2*A3+...+3^(n-1)*An=n/3,又A1+3A2+3^2*A3+...+3^(n-)*An-1=(n-1)/3,(比已知的式子最后少写一项,即有n-1项),两式相
1.证明:因为bn,a(n+1),b(n+1)成等比数列,所以[a(n+1)]²=bnxb(n+1)(n∈N*)a(n+1)=√[bnxb(n+1)]所以an=√[bnxb(n-1)](n≥
1、an,bn,a(n+1),所以,2bn=an+a(n+1)推出,2(bn+1)=a(n+1)+a(n+2)bn,a(n+1),b(n+1),所以,a(n+1)^2=bn*b(n+1),推出,a(n
(1)an=5Sn+1a(n-1)=5S(n-1)+1所以an-a(n-1)=5an,an=-a(n-1)/4,所以{an}是等比数列a1=5*a1+1,a1=-1/4所以an=(-1/4)^nbn=
令n=1得:a1=3.a1+2a2+2^2a3+…+2^(n-1)an=8-5n,把n换成n-1得:a1+2a2+2^2a3+…+2^(n-2)a(n-1)=8-5(n-1),相减得:2^(n-1)a
选Bn=1时a2=a1q>a1即a1q-a1>0a1*(q-1)>0a10所以q^(n-1)>0由于n为任意自然数所以q>0综上,答案选B,0再问:an+1=a1q^n>an=a1q^(n-1)怎么得
由已知an与1的等差中项等于Sn与1的等比中项得(an+1)/2=√SnSn=(an+1)²/4n=1时,S1=a1=(a1+1)²/4,整理,得(a1-1)²=0a1=
a1=2,a2=6,a3=10(an+2)/2=√2sn(an+2)^2=8sn(a(n-1)+2)^2=8s(n-1)相减:(an+2)^2-(a(n-1)+2)^2=8sn-8s(n-1)an^2
(1)用数学归纳法.A(n+1)=An^2-nAn+1=An(An-n)+1>=An*2+1>=(n+2)*2+1=2n+5>n+1+2(2)因为an>=n+2,所以an-n>=2A(n+1)=An(
∵an与1的差数中项等于√Sn∴a[n]+1=2√S[n]两边平方(a[n]+1)²=4S[n]①∴(a[n-1]+1)²=4S[n-1]②两式相减(a[n]+1)²-(