a是1到9的一个数字,求sn

1、Sn=(a1+an)n/2所以nan/Sn=2an/(a1+an)=2[a1+(n-1)d]/[2a1+(n-1)d]上下除以(n-1)=2[a1/(n-1)+d]/[2a1/(n-1)+d]n-

Sn=a+aa+aaa+...+aa..aaa(n个a)=a(1+11+111+11...111(n个1)=a/9*(10^1-1+10^2-1+10^3-1+...+10^n-1)=a/9*(10^

#includevoidmain(){longinti,j,a,n,Sn=0;scanf("%ld%ld",&a,&n);j=a;for(i=1;i

tn=a/9(10^i-1);这句话很多错误呀第一：C语言不支持10^i这样的你可以定义个函数求10的i次方第二：a/9(10^i-1);/在9与（10^i-1）之间要加*号第三：a/9*(10^i-

#includeusingnamespacestd;//Sn=a+aa+aaa+...+(n个a)//uA表示a//uN表示n//返回值为SnunsignedintSigmaN(unsignedint

#includemain(){intn;longa,sum=0;printf("pleaseinputaandn,andpressEntertocontinue\n");scanf("%ld%d",&

Sn=a*(1+11+111+1111+...+1111..11)=a*(n*1+(n-1)*10+(n-2)*100)+...1*10^(n-1))

#includevoidmain(){\x09intn;\x09longSn=0;\x09longt=1;\x09printf("pleaseinputn:\n");\x09scanf("%d",&n

给你个最简单的：#include"stdio.h"main(){doublen1,x,t,t1;intcx,i;scanf("%lf,%d",&n1,&cx);t=n1;t1=n1;for(i=1;i

#include#includevoidmain(){printf("\n请分别输入a和n,用逗号隔开:");intn=0;inta=0;scanf("%d,%d",&a,&n);intsum=a;f

设：x1=ax2=aaxn=aaaaa.a(n个)观察可得：x(n+1)=10xn+a等式两边同时加a/9得：x(n+1)+a/9=10xn+10a/9即：x(n+1)+a/9=10(xn+a/9)这

main(){inta,n,i,sum;scanf("%d%d",&a,&n);intsn[1000];sn[0]=a;sum=a;for(i=1;i

C#,C自己翻译一下就行了,算法是一样的(

#includevoidmain(){doublea,b,sn=0;inti,n;printf("pleaseinputa:");scanf("%lf",&a);printf("pleaseinput

#include"stdio.h"#include"math.h"voidmain(){inta,n,i;longsum=0;printf("Pleaseinputa(0

两边同时除以sn得1=1/2(sn-1)+1/sn设1/sn-a=-1/2(1/(sn-1)-a)解得a=2/3,又a1=2,所以1/s1-2/3=-1/6所以1/sn-2/3=（-1/6）（-1/2

int有范围限制2^32-1.超过了就成负数了.最高位为符号位.Sn改用longint吧

#includevoidmain(){inta,i,k,sum,n,t=0;scanf("%d",&n);sum=0,i=1,k=2;do{t=t+k;sum=sum+t;k=k*10;i++;}wh

c语言程序如下；#include#includevoidmain(){inti=0,n,k,a,sum=0;printf("请输入a,n的值:\n");scanf("%d,%d",&a,&n);whi

将循环小数转化为分数：4A6/999a/444=4A6/9999a=4A6*4显然4A6能被9整除,4+A+6必能被9整除,因为A是1到9中一个数字,A=8a=486*4/9=216