一个正整数n,计算s的前n项和
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设cn=an/2^n,求证cn是等差数列由cn=an/2^n,知an=2^n*cn,且a(n+1)=2^(n+1)*c(n+1),a(n-1)=2^(n-1)*c(n-1),由bn=2an-4a(n-
#includevoidmain(){floatsum=0.0;intn=0,sign=-1;scanf("%d",&n);for(floati=1;i
s=0;for(i=1;i
#includeintmain(){inti,k,n,t;doubleans;scanf("%d",&n);ans=0;t=1;k=1;for(i=1;i再问:输入输出示例Entern;3sum=0.
an+Sn=4096a(n+1)+S(n+1)=4096相减a(n+1)-an+a(n+1)=0a(n+1)/an=1/2所以是等比,q=1/2a1=S1所以2a1=4096a1=2048=2^11所
程序中的ab不知道做什么用的.pow(x,y)中x,y值反了,跟踪发现c=pow(x,-1)值为0这里因为c是整型,自动忽略小数位,值为0所以sum=sum+a*c;得到sum=0恒成立#includ
a(n)=2S(n-1)这时n的范围改变:n≥2两式作差,得:a(n+1)-a(n)=2a(n)所以公比为3,但是只符合a2以后的通项公式:当n=1时,a(n)=1当n≥2,且n∈N时,a(n)=2×
1)由S(n+1)=4an+2,知S(n)=4a(n-1)+2,两者相减,得S(n+1)-S(n)=a(n+1)=4[an-a(n-1)]由bn=a(n+1)-2an知,b(n-1)=an-2a(n-
/*1-2/3+3/5-4/7+5/9-6/11+.的前n项之和*/intmain(intargc,char*argv[]){intnum_n=1;inti=0;doubleresult=0.0;pr
intmain()doubleresult;inti;intn;scanf("%d\n",n);for(i=1;i
#includeusingnamespacestd;intmain(){inti;doublee,n,sum;cin>>n;for(i=0,e=-1,sum=0;i{e*=-1;sum+=e*1/(2
由于:{an}为等比数列则有:a(n)=a1q^(n-1),(a1,q不等于0)故:1/a(n)=(1/a1)(1/q)^(n-1)设首项a1=a则:其前n项之积M(n)=a(1)*a(2)*...*
#includeintmain(void){intn;inti;doublesum=0.0;intfact=1;scanf("%d",&n);for(i=1;i
a1=S1=a1S(n-1)=(n-1)^2*a(n-1)…………(1)Sn=n^2*an…………(2)(2)-(1)得an=n^2*an-(n-1)^2*a(n-1)当n>1时(n-1)^2*a(n
当n=1时,a1=-14;当n≥2时,an=Sn-Sn-1=-5an+5an-1+1,所以,又a1-1=-15≠0,所以数列{an-1}是等比数列;(2)由(1)知:,得,从而(nÎN*);
a1=s1=3设an=3+(n-1)d,则a2=3+d,a3=3+2ds3=9+3d=15所以d=2故an=2n+1,sn=(n+2)n
#includeintfib(intn){if(1==n||2==n)return1;returnfib(n-1)+fib(n-2);}intmain(){inti,n;scanf("%d",&n);
#includeintmain(){intn,i;doublesum=0;printf("Entern:");scanf("%d",&n);for(i=1;i
是要编程序么再问:C语言编程再答:#includeintjiecheng(intn){inti,mul=1;for(i=1;i