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等差数列前n项和例题

通项公式:等差数列an = a1+(n-1)d等比数列an = a1*q^(n-1)求和公式:等差数列前n项和Sn = n*a1 + n(n-1)/2 *d等比数列前n项和Sn = a1*(1-q^n)/(1-q) (q不等于1时)当q=1时,等比数列前n项和Sn = n*a1

(1)1+2+3+…+n=n(1+n)/2=n^2/2+n/2(2)1+3+5+…+(2n-1)=n(1+2n-1)/2=n^2(3)2+4+6+…+2n=n(2+2n)/2=n^2+n(4)1-2+3-4+5-6+…+(2n-1)-2n=(1-2)+(3-4)+(5-6)+…+[(2n-1)-2n]=-n

等差数列的前n项和Sn= na1+ n(n-1)d/2设An Bn 的首项 为 a b 公差分别为 c dAn/Bn= [2a+(n-1)c] /[2b+(n-1)d]An/Bn=(7n+4)/(n+3)待定系数法, 所以c =7d a =5.5d b =2dan=a+(n-1)c=5.5d +(n-1)c = 7nd-1.5dbn=b+(n-1)d=2d+(n-1)d=nd+da9/b9=an/bn= (14n-3):(2n-2)=(14n-14+11):(2n-2)=7+11:(2n-2)不可能是正整数0个

设首项为a,公差为d若公差为0,则S3/S6=1/2因此d≠0.由公式Sn=na+n(n-1)d/2得S3=3a+3dS6=6a+15dS12=12a+66dS3/S6=1/3(3a+3d)/(6a+15d)=1/3化简得a=2d∴S6=27d,S12=90d∴S6/S12=27/90=3/10选A

解: (1)a(n)=s(n)-S(n-1)=n^2+2n+4-[(n-1)^2+2(n-1)+4]=2n-1+2=2n+1(n≥2且n∈N*) 当n=1时,a(1)=S(1)=1^2+2+4=7(与通项公式不相符) a(1)=7,a(2)=5,a(3)=7 (2) n≥2时:a(n+1)-a(n)=2(n+1)+1-2n-1=2=d(定值) 故有:数列去除首项后所成的数列a(2),a(3),a(n)是等差数列

a(n)=a1+(n-1)d Sn=na1+n*(n-1)d/2 等差数列前N项和公式S=(A1+An)N/2 等差数列公式求和公式 Sn=n(a1+an)/2 或Sn=na1+n(n-1)d/2

an=a1+(n-1)*d;sn=(a1+an)*n/2; =(a1+a1+(n-1)d)*n/2; =a1*n+(n-1)*n*d/2; =5/6*n+(n-1)*n*(-1/6)/2=-5 5*2*n-(n-1)*n=-5*12 n^2-11n-60=0;(n-15)(n+4)=0;n=15,舍去(n=-4)n>0;

Sn=a1+a2+……+a(n-1)+an则由加法交换律Sn=an+a(n-1)+……+a2+a1相加2Sn=(a1+an)+[a2+a(n-1)]+……+[a(n-1)+a2]+(an+a1)因为等差数列中a1+an=a2+a(n-1)+……所以2S=n(a1+an)所以Sn=(a1+an)*n/2

1 an=3n-2 等差数列前n项和求和公式Sn=n*(a1+an)/2 a1=1 Sn=n(a1+an)/2=n(1+3n-2)/2=n(3n-1)/2 2 Sn=5n^2+3n an=Sn-S(n-1)=5n^2+3n-5(n-1)^2-3(n-1)=10n-2 所以a1=10-2=8 a2=20-2=18 a3=30-2=28

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