求斐波那契函数前20项之和

为用了很没有效率的递归,所以出结果有点慢#includeiostream.h

帮你写好了.unsigned int fib(unsigned int n) {\x09if (n == 1 

根据公式Sn=na+[n(n-1)d/2]所以：20=10a+10(10-1)d/260=20a+20(20-1)d/2a=1.1,d=0.2S30=30×1.30×(30-1)×0.2/2=120

因为用了很没有效率的递归,所以出结果有点慢#includef(int);main(){inti,s=0;for(i=1;i

main(){inti,n,s=1,f[]={0,1,1};printf("Pleaseinputthenumberofterms:");scanf("%d",&n);if(n==0){s=0;f[2

//分别使用两个递归求分子分母即可：代码如下：usingSystem;namespace数列求和{classProgram{staticvoidMain(string[]args){intresult

通项公式为A（n-1）+An=A(n+1)11235813213455＝143

143,绝对正确

#include#defineCOL5//一行输出5个longfibonacci(intn){//fibonacci函数的递归函数if(0==n||1==n){//fibonacci函数递归的出口re

// C++int F(int n) {if (n == 0) return 1;else if

1112233455861372183495510891114412233133771461015987161597172584184181196765201094621177112228657234

1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,10946,17711,28657,46368,75025,121

PrivateFunctionbq(ByValsAsLong)AsLongSelectCasesCase1bq=1Case2bq=1CaseIs>=3bq=bq(s-1)+bq(s-2)EndSele

#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]

不用那么麻烦inta=1,b=2,i,k,n;floatsum=0.0;scanf("%d",&n);for(i=0;i再问：不是题目要用递归函数

由公式Sn=na1+n(n-1)d/2有Sp=pa1+p(p-1)d/2=q.(1)Sq=qa1+q(q-1)d/2=p.(2)(1)-(2)得(p-q)a1+(p+q-1)(p-q)d/2=q-p∵

 Private Sub Command1_Click()Dim F(11), i As LongF(0) = 

#includevoidfib(intn,intf0,intf1){intf;//当前项inti=0;if(n=2)printf("%8d,%8d",f0,f1);//f0,f1for(i=2;i

PrivateFunctionbq(ByValsAsLong)AsLongSelectCasesCase1bq=1Case2bq=1CaseIs>=3bq=bq(s-1)+bq(s-2)EndSele

设等比数列的首项为a1,公差为d,则由已知得,a1+a1q+a1q^2+a1q^3=a1(1+q+q^2+q^3)=2⑴a1+a2+a3+...+a8=a1(1+q+q^2+...+q^7)=6⑵⑵/