VB题目求数列2 1 3 2

dimf()asdoublen=inputbox("in","NO.")redimf(n)asdoublef(1)=1f(2)=1fori=3tonf(i)=f(i-1)+f(i-2)nextprin

PrivateSubForm_Click()Dima(1To40)AsLongFori=1To40Ifi=1Ori=2Thena(i)=1Elsea(i)=a(i-2)+a(i-1)EndIfPrin

#include#includemain(void){intn,i;printf("请输入n的值\n");scanf("%d",&n);intsum1=1,sum2=2,sum;for(i=3;i

第一问：因为Sn=n^2+3n所以S1=a1=1^2+3*1=4=b1则Sn=na1+n(n-1)d/2=n^2+3n,化简可得出d=2a31=a1+30d=4+30*2=64b3=4q^2=64所以

DimnAsIntegerDimmAsIntegerDimiAsIntegerDimjAsIntegerDima(30)AsInteger'确定数组的范围（）里的数即为最大可储存数+1PrivateS

PrivateSubForm_Click()DimnAsIntegern=Val(InputBox("请输入N:"))Dima,bAsLonga=1:b=1Fori=1TonPrinta&""&b&"

OptionExplicitDimf(40)AsLongPrivateSubCommand1_Click()DimiAsByteDimsAsLongf(1)=1f(2)=1s=2Print"No1:"

非递归：staticvoidf(intn){longp1=1,p2=1,p=1;for(inti=1;i

(m-1)+2=m+1所以am-1+a2=am+a1=8等差数列{an},a1,a2,a3,…,am的和为64即：m(a1+am)/2=64所以8m/2=4m=64所以m=64/4=16项数m=16.

dima,b,c,x1,x2asinteger点计算之后a=textboxA.textb=textboxB.textc=textboxC.textx1=(-b-(b^2-4*a*c)^0.5)/(2*

1填写imod6=0,指的是余数要等于0；2填写S,表示要把S输出.

/>n=1时,a1=1/2n≥2时,Sn=n²×anS(n-1)=(n-1)²×a(n-1)an=Sn-S(n-1)=n²×an-(n-1)²×a(n-1)(n

我的理解是：{an+an-1}是公比为3的等比数列,{an-3an-1}是公比为-1的等比数列,这样就该知道了吧~等比数列中an=a1乘qn-1次方=a2乘qn-2次方,an-1=a1乘qn-2次方,

楼上的程序会慢死人的.给一个非递归实现.functionFibonacci(byvalnasLong)asLongdiml1aslong,l2aslong,l3aslongl1=1l2=1ifn

解题思路：考察二项展开式的通项解题过程：varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/re

看不清楚,不然可以帮你.再问：谢谢啦。再答：4.(1)当n=1时，a1=S1=(a1-1)/3,∴a1=-1/2n=2,a1+a2=S2=(a2-1)/3,∴a2=1/4,(2)当n≥2时，a(n)=

1.an+2a(n+1)+a(n+2)=0即a1q^(n-1)+2a1q^n+a1q^(n+1)=0∵a1≠0,q≠0,两边同除以a1q^(n-1),得1+2q+q^2=0所以q=-1所以S101=a

先写两个数列.PublicSubaa()DimabAsStringDimi,jAsIntegerDima,bAsInteger'素数序列Fori=1To1000Forj=1ToiIfiModj=0An

B.A(1)=1*1=1,A(2)=2*2=4,...,A(5)=5*5=25C.for循环结束后m为10~1递减数列,m(2+m(x))=m(2+m(6))=m(2+5)=m(7)=4

A1=3*1^2-1+2A2=3*2^2-2+2..An=3*n^2-n+2迭加得∑An=3∑n^2-∑n+2n=2n(n+1)(2n+1)-n(n+1)/2-2n