fib(n)要求计算此数列的前20项

设通项为anSn-Sn-1=an=2^(n-1)(n≥2)又S1=a1=1符合条件,故an=2^(n-1)(n∈N*)于是奇数项的前n项和NN=a1+a3+...+a2n-1=1+2^2+2^4+..

//Author:Yach//QQ:79564727//申请请注明来者何人..#includeusingnamespacestd;intfib(int);voidmAndn(int,int);intf

Sn=2^n-1sn-1=2^(n-1)-1an=sn-sn-1=2^n-2^(n-1)=2^(n-1)当n＝1a1＝1也成立,所以an=2^(n-1)即｛an｝是公比为2的等比数列,首项为1所以此数

用VB写的~PrivateSubForm_Load()Dimfn&,f1&,f2&,n%f1=1:f2=1n=Val(InputBox("请输入n"))fn=f1+f2Printf1Printf2Pr

#includeintmain(){inti=0;floatsum=0;intn;intx[n],y[n];printf("请输出计算的项数:");scanf("%d",&n);x[0]=2;x[1]

∵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

#include"stdio.h"voidmain(){intn;inta,b,c;scanf("%d",&n);a=0;b=1;if(n==1)printf("0\n");elseif(

intfib(n){if(n

Sn=2^n-1---------(1)当n=1时,a1=1S(n-1)=2^(n-1)-1-------(2)(1)-(2)Sn-S(n-1)=2^n-2^(n-1)an=2^(n-1)a1+a3+

#include"stdio.h"#include"math.h"intmain(void){inti,m,n;intrepeat,ri;longf;longfib(intn);inta,b,c;sc

PrivateSubCommand1_Click()Dima(30)asIntegerIfOption1.Value=TrueThena(1)=1a(2)=1Fori=3To30a(i)=a(i-1)

An=Sn-Sn-1=2^n-2^(n-1)=2^(n-1)此数列奇数项的前n项为等比为2^2=4,其和为：A1+A3+...+A(2n-1)=1+2^2+...+2^(2n-2)=（1-2^n)/1

你能再发一张清楚些的照片吗?再问：抱歉我没办法

Sn=(1/6)n(n+1)(2n+1)用数学归纳法证当n=1时,S1=a1=1,成立假设n=k时成立,则n=k+1时Sn+1=Sn+(n+1)2=(1/6)n(n+1)(2n+1)+(n+1)2=(

1sn=2^n-1sn-1=2^(n-1)-1an=sn-sn-1=2^n-2^(n-1)=2^(n-1)a1=s1=2^1-1=1也满足an所以an=2^(n-1)所以an是首项为1,公比为2的等比

当n=1时，a1=S1=21-1=1，当n≥2时，an=Sn-Sn-1=2n-1-（2n-1-1）=2•2n-1-2n-1=2n-1，对n=1也适合∴an=2n-1，∴数列{an}是等比数列，此数列奇

错项相减法

因为(n+1)^3-n^3=(n+1-n)[(n+1)^2+n(n+1)+n^2]=3n^2+3n+1所以3n^2=(n+1)^3-n^3-3n-1所以3*1^2+3*2^2+……+3n^2=[(1+

#includevoidmain(){intn;longf1,f2,s;coutn;f1=1;f2=1;s=0;if(n==0)cout

Sn=n^2-10na1=S1=1^2-10*1=-9当n≥2时an=Sn-S(n-1)=n^2-10n-[(n-1)^2-10(n-1)]=2n-11把a1代入发现也符合通项公式所以an=2n-11