计算1 3! 5! ... (2n-1)! 前10 项和.

intaddsum(intm,intn){intm,ncin>>m>>n;if(m<n)cout<<"错误,m<n"<<endl;intsu

PrivateSubCommand1_Click()Dimn%,i%,y%,a%a=0y=1n=Val(text1.Text)Fori=1Tona=a+1y=y*aNextiEndSu

ak=kC(k,n)=k*n!/k!*(n-k)!=n*(n-1)!/(k-1)!(n-k)!=nC(k-1,n-1)故原式=nC(0,n-1)+nC(1,n-1)+nC(2,n-1)+……+nC(n

101+103+105+107+.+(2n-5)+(2n-3)+(2n-1)+(2n+1)=（101+2n+1)×（n+1-50)/2=(n+51)(n-49)

用夹逼定理：(6^n)^(1/n)≤(2^n+3^n+4^n+5^n+6^n)^(1/n)≤[5倍的(6^n)]^(1/n)三边同时取极限,第一项(无论是否取极限)永远恒等于6,中间就是要求的极限,右

原式=[n(n+3)[(n+1)(n+2)]+1=(n2+3n)[(n2+3n)+2]+1(n2+3n)2+2(n2+3n)+1=(n2+3n+1)2=n2+3n+1．

S=1!+2!+3!+…+n!,得n!=n*(n-1)*(n-2)*…*2*1

PrivateSubForm_Click()Dimi,n,tAsIntegerDimsAsSingles=0t=1n=InputBox("n=")Fori=1Ton-1s=s+t*i/(i+1)t=-

设A=1+3+5+7…+（2n+1）①,所以A=（2n+1）+(2n-1)+(2n-3)+……+1②①+②=[1+(2n+1)]+[3+(2n-1)]+……+[(2n+1)+1]=(2n+2)+(2n

(1×2×3+2×4×6+…+n×2n×3n)÷(1×3×5+2×6×10+…+n×3n×5n)=1*2*3(1+2+.+n)÷1*3*5(1+2+...+n)=2/5

如图所示

1+3+5+…+(2n+1)＝(2n+2)(n+1)/2=(n+1)^2原式子化为lim[(1+1/n)^n]^2而lim[(1+1/n)^n]^＝e当然答案就是e^2

1+(2n-1)=2n3+(2n-3)=2n5+(2n-5)=2n……2n-1+1=2n2s=2n+2n+……+2n（n个）2s=2n^2s=n^2

1计算：1+3+5+…+（2n+1）=(n+1)^22+4+6+…+2n=n(n+1)1+3+5+…+（2n+3）=(n+2)^21+4+7+…+（3n+1）=(n+1)(3n+2)/22.等差数列1

2+4+6+……+2n=n(n+1)1+3+5+……+(2n+1)=(n+1)^21+2+3+……+2n+(2n+1)=(2n+1)(n+1)

(1+n)*n/2适用于等差数列：(首项+末项)*项数/2=数列和例题：1+2+3+4+5……+99+1001就是首项,100就是末项,一共有100个项数1+2+3+...+100=(1+100)*1

DATASEGMENTNDB2SDB1DATAENDSCODESEGMENTASSUMECS:CODE,DS:DATASTART:MOVAX,DATAMOVDS,AXS:MOVAL,NI

√(n^2+4n+5)－(n-1)＝[(n^2+4n+5)－(n-1)^2]/[√(n^2+4n+5)＋(n-1)]＝(6n+6)/[√(n^2+4n+5)＋(n-1)]＝(6＋6/n)/[√(1＋4

9^n*(1/27)^n+1*3^n+2=（9/27）^n+3^n+2=1/(3^n)+3^n+2=[3^(-n/2)+3^(n/2)]^2