已知正数项列an的前n项和为sn,根号下sn是四分之一与
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![已知正数项列an的前n项和为sn,根号下sn是四分之一与](/uploads/image/f/4272396-60-6.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E6%95%B0%E9%A1%B9%E5%88%97an%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BAsn%2C%E6%A0%B9%E5%8F%B7%E4%B8%8Bsn%E6%98%AF%E5%9B%9B%E5%88%86%E4%B9%8B%E4%B8%80%E4%B8%8E)
当n=1时,2S1=a1+1/a1,得a1=1当n=2时,2S2=2(1+a2)=a2+1/a2,得a2=√2-1当n=3时,2S3=2(√2+a3)=a3+1/a3,得a3=√3-√2猜想an=√n
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证明:∵Sn=an(an+1)2∴S1=a1(1+a1)2∴a1=1…(1分)由2Sn=a2n+an2Sn-1=a2n-1+an-1⇒2an=2(Sn-Sn-1)=a2n-a2n-1+an-an-1…
当k-1≤x≤k时,有1/k^(2010/2009)≤1/x^(2010/2009)1/k^(2010/2009)=(k,k-1)∫1/k^(2010/2009)dx≤(k,k-1)∫1/x^(201
Sn是an^2和an的等差中项所以Sn=(an²+an)/2①同理得Sn-1=(an-1²+an-1)/2②①-②得2an=an²-an-1²+an-an-1化
1楼貌似错了!(a1^2-3a1=6a1与An^2+3An=6Sn矛盾)An^2+3An=6SnA(n+1)^2+3A(n+1)=6S(n+1)后减前得A(n+1)^2+3A(n+1)-An^2-3A
2√Sn=an+1则有,4Sn=(an+1)²4a(n+1)=4[S(n+1)-Sn]=[a(n+1)+1]²-(an+1)²=[a(n+1)]²+2a(n+1
(1)(an+2)/2=根号下2Sn所以8Sn=(an+2)^2n=1,S1=a1.8a1=(a1+2)^2,得a1=2n=2,8S2=(a2+2)^2,8(a1+a2)=(a2+2)^2,得a2=6
Sn=n^2推出an=2n-1bn=(2n-1)/3^nTn=b1+b2+b3+……+bn-1+bn=1/3+3/3^2+5/3^3+……+(2n-3)/3^n-1+(2n-1)/3^n①3Tn=1+
1.n=1时,2a1=2S1=a1²+1-4a1²-2a1-3=0(a1+1)(a1-3)=0a1=-1(数列各项均为正,舍去)或a1=3n≥2时,2an=2Sn-2S(n-1)=
由a1^3+a2^3+a3^3+.+an^3=Sn^2得a1^3+a2^3+a3^3+.+an^3+an+1^3=Sn+1^2两式相减得:an+1^3=Sn+1^2-Sn^2=(Sn+1+Sn)(Sn
(1)为了不造成混乱记a=A(^)表示多少次方S1=1/4(A1+1)^2得A1=1Sn=1/4(An+1)^2Sn-1=1/4(An-1+1)^2两式相减得An=1/4[(An+1)^2-(An-1
1)n=1,解得a1=1n>1时S(n-1)=1/4(a(n-1)+1)^2Sn=1/4(an+1)^2相减并整理得到an^2-2an-a(n-1)^2-2a(n-1)=0(an-a(n-1)-2)(
根据A1=S1(n=1);An=Sn-Sn-1(n>=2)可得An=2n-1;进而得Bn=(2n-1)/3^n下证Tn=1-(n+1)/3^n显然T1=1/3=B1Tn-Tn-1=1-(n+1)/3^
sn=(1/8)(an+2)²S(n-1)=(1/8)[a(n-1)+2]²an=Sn-S(n-1)=(1/8){(an+2)²-[a(n-1)+2]²}=(1
由题意得1S3=a1+a2+a3=7……1;6a2=a1+1+a3+6……22式+1式得a2=2……3将3式代入12得q=2或1/2a1=4或1an=4*(1/2)^(n-1)或an=2^(n-1)2
(1)a1=(a1+1)24,解得a1=1,当n≥2时,由an=Sn-Sn-1=(an+1)2−(an−1+1)24,得(an-an-1-2)(an+an-1)=0,又an>0,所以an-an-1=2
Sn=n^2S(n-1)=(n-1)^2an=Sn-S(n-1)=n^2-(n-1)^2=2n-1因此得到数列{an}的通项公式为an=2n-1
2*Sn^(1/2)=An+1(1)2*S1^(1/2)=A1+1,S1=A1A1=1(2)Sn=(An+1)^2/4S(n-1)=[A(n-1)+1]^2/4An=Sn-S(n-1)=(1/4)*(
由题意知2an=Sn+1/2,an>0,当n=1时,2a1=a1+1/2,解得a1=1/2,当n≥2时,Sn=2an-1/2,S(n-1)=2a(n-1)-1/2,两式相减得an=Sn-S(n-1)=