已知数列an为等差数列 sn为其前n项和,2sn=an 2n
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1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-
缺少条件,{an}为正项数列,否则log3(an)无意义,题目没法解.证:数列为正项数列,公比q>0a(n+1)/an=qb(n+1)-bn=log3[a(n+1)]-log3(an)=log3[a(
(1)由已知,n,an,Sn成等差数列,所以Sn=2an-n,Sn-1=2an-1-(n-1),(n≥2)两式相减得an=Sn-Sn-1=2an-2an-1-1,即an=2an-1+1,两边加上1,得
已知:数列an满足a1=2,其前n项和为Sn=n+7-3an;数列bn满足bn=an-1,证明数列bn是等差数列.代入an=Sn-S(n-1),得Sn=n+7-3(Sn-S(n-1)),变形成:Sn-
数列{Sn/n}构成一个公差为2的等差数列,∴Sn/n=2n,∴Sn=2n^2,∴a3=S3-S2=18-8=10.
根据题意:首项为a1,公差为d3a8=5a13因为:a8=a1+7da13=a1+12d所以:3(a1+7d)=5(a1+12d)3a1+21d=5a1+60da1=-19.5d即:a20=a1+19
a5=a1+4d=8S5=(a1+a5)*5/2=5a1+10d=20d=2a1=0an=2*(n-1)Sn=(a1+an)*n/2=(n-1)*n(2)令bn=Sn+2an+1/6=n^2+3n-4
a3=a1+2d=6S3=a1+a2+a3=3a1+3d=12解得a1=2,d=2,故an=2n所以Sn=n(n+1)所以1/S1+1/S2+……+1/Sn=1/(1*2)+1/(2*3)+1/(3*
a1+2d=11(a1+a1+8d)*9/2=153∴a1=5d=3∴an=5+3(n-1)=3n+2
S(2n-1)=(2n-1)an,T(2n-1)=(2n-1)an,所以an/bn=S(2n-1)/T(2n-1),所以a9/b9=S17/T17=18/31.
设首项为a1,方差为da1=a3-2d=11-2d,a9=a3+6d=11+6dS9=n(a1+a9)/2=9*(11-2d+11+6d)/2=153d=3a1=a3-2d=11-2d=5通项公式=a
s9=9a1+9×8÷2×d=1539a1+36d=153a1+4d=17a1+2d=11所以a1=5d=3所以an=a1+(n-1)d=5+3(n-1)=3n+2
a3=-13(a1+a9)*9/2=-45a1+a9=-10所以a1+2d=-13,2a1+8d=-10所以a1+2d=-13,a1+4d=-5解得d=4a1=-21an=-21+4(n-1)=4n-
解:①当n=1时a1=S1=2②当n≥2时an=Sn-Sn-1Sn=3n^2-nSn-1=3(n-1)²-(n-1)所以an=6n-4=2+6(n-1)带入n=1得到a1=2符合①综上所述a
∵{log2an}是公差为-1的等差数列∴log2an=log2a1-n+1∴an=2log2a1−n+1=a1•2−n+1∴S6=a1(1+12+…+132)=a1•1−1261−12=38,∴a1
(1)设首项和公差分别为a1,d由a3=7S4=24得a1+2d=74a1+6d=24所以a1=3d=2,则an=2n+1;(2)2Sp+q-(S2p+S2q)=2(p+q)2+4(p+q)-4p2-
设an=a1+(n-1)d=10+(n-1)dSn=na1+(n-1)nd/2=10n+(n-1)nd/2S12=120+66d=-125那么d就算出来了d=-245/66所以an=10+(n-1)(
当n≥2时,可以化为Sn-S(n-1)=-2Sn×S(n-1),两边同除以Sn×S(n-1),得1/Sn-1/S(n-1)=2所以{1/Sn}是以2为首项,2为公差的等差数列即1/Sn=2nSn=1/
因为Sn=3n^2+5nS(n-1)=3(n-1)^2+5(n-1)两式相减所以an=6n-3+5=6n+2所以an=8+6(n-1),所以an是以8为第一项,公差为6的等差数列.