函数sum1计算1 3 -n的和,当n为偶数时,调用函-搜答案-智慧作业帮

#include <stdio.h>int main(void){ int n; &nbs

#include#includedoublefun(intn);intmain(){printf("Inputn:");intn;scanf("%d",&n);doubleS=fun(n);

（1）令S(x)=∑(n=0→无穷）n*x^n/(n+1)则S(x)=x/2+2/3*x^2+3/4*x^3+···+n/(n+1)*x^n+···(1)两边同乘x：xS(x)=1/2*x^2+2/3

#includemain(){\x05intm,n;\x05ints=1;\x05intpower(inta,intb);\x05printf("inputmn(m^n):\n");\x05scanf

intfunc(intx,intn){inty;if(n

你确定函数类型是要double?这个是整型的#includeintfact(int);voidmain(){intn;scanf("%d",&n);for(inti=1;i

floatfun(intn){floatsum=0;inti;for(i=1;i

参考一下我的：OptionExplicitPrivateSubCommand1_Click() DimiAsInteger,nAsInteger,nRandAsInteger&n

intsun=1;//计算阶乘的值intsum=0;//累加for(inti=0;i

∑[x^2n\(2n-1)]=x∑[x^(2n-1)\(2n-1)](把x提出来了)设g(x)=∑[x^(2n-1)\(2n-1)]一阶导数g'(x)=∑x^(2n-2)=1/(1-x^2)g(x)=

#includeintmain(){intn;doublem;doublefac(intn);scanf("%d",&n);m=fac(n);printf("%d!=%f\n",n,m);return

Subshiyan()DimInputValue,ReturnValueAsIntegerInputValue=InputBox("请输入一个自然数","实验",1)IfInputValueMsgBo

源码如下：#includeintsum(inta,intb){\x09returna+b;\x09}intmain(){\x09intm=1,n=2,k=3;\x09intt=sum(sum(m,n)

publicinttotal(intn){intresult=0;for(inti=1;i

#includeusingnamespacestd;longf(intn);intmain(){intm;longsum=0;cin>>m;for(inti=1;i

if(n==1)\x05\x05return1;你可以改成n==0或者改成returnx取其中一种就可以了

#includeusingnamespacestd;voidmain(){inta=0,b=0;cin>>a>>b;cout

#includevoidmain(){inta,b,sum=0;printf("请输入两个整数：");//将两个改成n个就好了scanf("%d%d",&a,&b);sum=a+b;printf("%

1/(1-x^2)=∑(1到+∞)x^(2n-2)(-1

intfun(intn){inti,jiech;jiech=1;for(i=1;i