求数列,前两项分别为1和2,其余各项为其前两项之和.求该数列前20项之和.
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floatfunction(intn){floatave,a[100],sum=0;inti;if(n==1)return1;elseif(n==2)return1.5;elseif
因为An+1=2SnAn=2S(n-1)所以A(n+1)-An=2AnA(n+1)/An=3是公比为3,首项a1=1的等比数列,An=A1*q^(n-1)即An=3^(n-1)
an=a1+(n-1)d=2n+1Sn=[n(a1+an)]/2=n(n+2)1/Sn=1/n(n+2)=[1/n-1/(n+2)]/2An=S1+S2+...+Sn=21/40-(2n+3)/2(n
functionsum(nasinteger)dima(30)asintegerdimsasintegers=0a(1)=0a(2)=0a(3)=1fori=4to30a(i)=a(i-1)+a(i-
#include"math.h"main(){inta[30],sum,i=3;a[0]=0;a[1]=0;a[2]=1;sum=a[0]+a[1]+a[2];while(i
n=b^2n,Tn=b^2+b^4+b^6+……+b^2n=b^2n(1-b^2n)/(1-b^2)所以1-bn=1-b^2n所以(1-bn)/Tn=(1-b^2n)/{b^2(1-b^2n)/(1-
1.S3=3a3-3d=12,a3-d=4d=2a2=4,a1=2an=2n2.Sn=2n+n(n-1)=n²+n=n(n+1)1/Sn=1/n-1/(n+1)1/S1+1/S2+·····
/>错位相减求和Sn=1/2^1+3/2^2+5/2^3+.+(2n-3)/2^(n-1)+(2n-1)/2^n①‘①×1/2(1/2)Sn=1/2^2+3/2^3+.+(2n-3)/2^n+(2n-
(1)2Sn=an^2+an2Sn-1=a(n-1)^2+a(n-1)2an=2Sn-2Sn-1=an^2-a(n-1)^2+an-a(n-1)an^2-a(n-1)^2=an+a(n-1)[an+a
#includeintmain(){intf[20]={1,1},i;for(i=2;i再问:c语言:一组数组,十个学生的成绩,输出最高分和最低分(可以再帮帮忙不)再答:#includeintmain
因为:An+1=2Sn,则A(n-1)+1=2S(n-1)那么:2Sn-2S(n-1)=(An+1)-(A(n-1)+1)(n>=2)又因为:2Sn-2S(n-1)=2An(n>=2)所以:2An=(
sn=2an-1s(n+1)=2a(n+1)-1a(n+1)=s(n+1)-Sn=2a(n+1)-2an得a(n+1)/an=2所以数列{an}是公比为2的等比数列,a1=s1=2a1-1,a1=1a
1因为Sn=3/2an-1/2所以S(n-1)=3/2a(n-1)-1/2两式相减得:an=3/2(an-a(n-1))化简得:an=3a(n-1)当n=1时,S1=3/2a1-1/2,S1=a1,解
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
七项为a1a2a3a4a5a6a7,奇数项a1+a3+a5+a7=24,a3+a5=12,a5=a3+2,a3=5,得到a1=3,a3=5,a5=7,a7=9偶数项a2*a4*a6=a4^3=64,a
令n=1,得4a1-2a1=1,解得a1=1/2当n≥2时,4an-2Sn=1①4a(n-1)-2S(n-1)=1②①-②得4an-4a(n-1)-2an=0即2an=4a(n-1)所以an/a(n-
设第一个数为a1a1(1+q+q^2)=14a1^3*q^3=64q=2或1/2a1=2或8这个数列为{8,4,2}或{2,4,8}清补全第二问
an=1+2n,sn=(a1+an)/2n=(3+1+2n)/2*n=n(n+2),1/sn=1/n(n+2)=1/2*(1/n-1/(n+2)),Tn=1/2{1+1/2-1/(n+1)-1/(n+
设第N(N>3)项数值为x,则x=((N-1)+(N-2)+(N-3))/2,令x>1200,则((N-1)+(N-2)+(N-3))/2〉1200,N〉802,故该数列从第803项开始,其数值超过1
an=a1+(n-1)d=2+(n-1)*1=n+1Sn=a1n+n(n-1)*d/2=2n+n(n-1)/2=2n+n^2/2-n/2=(n^2+3n)/2