写一个函数,求下列前N项之和

floatfunction(intn){floatave,a[100],sum=0;inti;if(n==1)return1;elseif(n==2)return1.5;elseif

因为an=（5/9）×((10^n)-1))所以sn=（5/9）×((10^1)-1))+（5/9）×((10^2-1))+…………+（5/9）×((10^n)-1))=（5/9）×[(10^1+10

an=(5/9)(10^n-1)接下来等比数列求和an=(5/9)10^n-(5/9)bn=an+(5/9)=(5/9)10^nSbn=(50*10^n-50)/81=San+(5/9)nSan=(5

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

#includeintmain(){inti=1,j=1,n;floatsum=0;scanf("%d",&n);for(intk=0;k

设该等差数列的首项为a1，公差为d，由已知得10a1+10×9d2＝31020a1+20×19d2＝1220，解得a1＝4d＝6．∴此数列的前n项和公式为Sn＝4n+n(n−1)2×6＝3n2+n．

PrivateSubCommand1_Click()n=Val(Text1.Text)Fori=1Tons=s+iNextiText2.Text=sEndSu

#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=(103n-3n^2)/2S1=a1=50Sn-1=[103(n-1)-3(n-1)^2]/2Sn-Sn-1=an=53-3na1a2……a17都是正数,后面的是负数设Tn=｜an｜的n项之和n

#includemain(){inti,j,n,k,t;doublesum=2;printf("inputanumber:\n");scanf("%d",&n);j=2;k=1;for(i=1;i再问

#includeusingnamespacestd;voidfan(intn,inta[][100]);voidmain(){intb[100][100];inti,j;intm;cout

如果是要求前N个偶数之和：#includeusingnamespacestd;longFunctionadd(intn){longsum=0;for(inti=0;i

importjava.util.Scanner;publicclassOushu{Oushu(){try{System.out.println("请输入n值");Scannerin=newScanne

设等比数列首项为a,比为q则Sn=a(1-q^n)/(1-q)倒数的数列首项为1/a,比为1/qTn={1/a[(1-(1/q)^n)]}/(1-1/q)=q(q^n-1)/[aq^n(q-1)]Sn

#include#include#defineN3voidgetDiagonalValue(inta[N][N]){inti;intsum1,sum2;sum1=sum2=0;for(i=0;i

设A(n)=1+3×2+5×2²+7×2³+...+(2n-1)*2^(n-1)2A(n)=2+3×2²+5×2³+...+(2n-3)*2^(n-1)+(2n

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

已知数列{an}是等差数列.（1）若前四项和为21,后四项之和为67,前n项和为286,求项数n（2）若Sn=20,S2n=38,求S3n（1）a1+a2+a3+a4=21an+an-1+an-2+a

由等差数列的定义和性质可得首项与末项之和等于21+674=22，再由前n项和为286=n(a1+an)2=11n，n=26，故选B．

错位相减法,Sn=(-1)^n*(3n/2-1/4)-1/4