已知数列an是等差数列 其前n项和公式为sn,a3=1,s3=-3
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证明an=Sn-S(n-1)=100n-n^2-[100(n-1)-(n-1)^2]=100n-n^2-[100n-100-(n^2-2n+1)]=100n-n^2-(-n^2+102n-101)=1
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-
因为Sn-Sn-1=n^2-3n-{(n-1)^2-3(n-1)}=2n-4.又由an=Sn-Sn-1,所以an=2n-4,最后还要验证一下,当n=1时,S1=a1,符合题意.d=an-an-1=2易
an=sn-s(n-1)=13-2n(n>1)a1=s1=11所以an=13-2n(n>0)当n>1,有an-a(n-1)=-2所以an是等差数列再问:(2)求数列﹛|an|﹜前n项的和。再答:前n项
∵sn=2n^2-n∴s(n-1)=2(n-1)^2-(n-1)=2n^2-3n+3当n≥2时,sn-s(n-1)=an=(2n^2-3n+3)-(2n^2-n)=-2n+3(n>1)当n=1时,a1
1、a3=6,S3=12所以a1+2d=6,3a1+3d=12所以a1=2,d=2所以{an}的通项公式为an=2+(n-1)*2=2n2、Sn=2n+n(n-1)=n²+n=n(n+1)1
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*
由S3=S7,则S7-S3=a4+a5+a6+a7=0,ak=a1+(k-1)d,d为方差,可得2a1+9d=0,已知a1=-9,则d=2.求sn的最小值即要看数列从第几项开始非负,可知a5=-1,a
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-
a3=a1+2d=11s9=[9(a1+a9)]/2=153所以a1+a9=34(a1+a1+8d=34)然后a1+2d=112a1+8d=34(a1+4d=17)方程组解得a1=5;d=3an=a1
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
(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-
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.
证::n=1,a1=s1=4n>1an=Sn-Sn-1Sn=n^2+3nSn-1=(n-1)^2+3(n-1)an=2n+2经验证n=1满足通项n>1an-an-1=2,由等差数列定义可知,数列{an
Sn=na1+[n(n-1)/2]dS5=5a1+10ds10=10a1+45dS15=15a1+105dS10-S5=5a1+35dS15-S10=5a1+60d(S10-S5)-S5=25d(S1
因为S(6)=6a(1)+6×5d/2=6a(1)+15dS(12)=12a(1)+66dS(18)=18a(1)+153d而S(12)-S(6)=6a(1)+51dS(18)-S(12)=6a(1)
an=a1+(n-1)d;Sn=(a1+an)*n/2S6=3(a1+a6);①S12-S6=3a1+6a12-3a6②S18-S12=3a1+9a18-6a12③②-①=6a12-6a6=6(a1+