求前n项和的方法 C语言中怎么输入N

设数列{an}是等差数列,其公差为d,d≠0,根据等差数列的定义：an-a(n-1)=d∴a2-a1=da3-a2=da4-a3=d.an-a(n-1)=d上述各式相加：an-a1=(n-1)d即：a

是前n个元素么?main(){inti=n,sum=0;max=a[0];min=a[0];for(i=0;ia[i])?max:a[i];min=(min再问：输出的avg不是整数再答：哎呀，忘了这

#defineN10;main(){inti,j;inta[N][N];intsum=0;for(i=0;i

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

解题思路：该题利用错位相减法求数列的前n项和，这种题型方法固定，但是计算量大，所以借助补0来让项错位，防止出错解题过程：

#includeintmain(){intn,i,t;floats,a;scanf("%d",&n);a=0;s=0;for(t=n;t>=1;t--){a=0;for(i=1;

#includeintmain(){inti,sum=0;for(i=1;i

main(){inta,sum,i;for(i>0;i

#includeintf(intn){if(n==1)return1;elseif(n==2)return2;elsereturnf(n-1)+f(n-2);}intmain(){intn;print

#include#defineCOL10//一行输出10个longscan(){//输入求fibonacci函数的第N项intn;printf("InputtheN=");scanf("%d",&n)

#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;i

#include#definemaxsize50main(){intm,n,d,i,count;intA[maxsize];\x09printf("\n请输入n,m的值,以逗号分开:");\x09sc

#include#include#defineN200longintF[N];voidFi(intn);longintSum(intn);voidmain(){intn;longintsum;prin

a1=2a2-a1=3*2^(2-1)=6令cn=a(n+1)-an=3*2^(2n-1),则c1=a2-a1=6,cn/c(n-1)=4cn是首项是6公比是4的等比数列设cn的前n-1项和为s(n-

#includeintFib(intm){if(m==1||m==2)return1;returnFib(m-1)+Fib(m-2);}voidmain(){intn,i;printf("请输入n的值

S(n)=1*2+3*2^2+5*2^3+.+(2n-3)*2^(n-1)+(2n-1)*2^n-----------------------(1)2S(n)=1*2^2+3*2^3+.+(2n-3)

#includevoidmain(){intn,i=1;doublea=0,x;printf("请输入要求X的前几项之和：");scanf("%d",&n);printf("请输入x的前m项(m>=n

#includevoidmain(){longintf1,f2;/*定义为长整型,后面的数很大,整型容纳不下*/inti,n;f1=1;f2=1;printf("inputn:");scanf("%d

解题思路：主要利用分组求和和等差数列求和公式，即可求出100项的和解题过程：