计算数列1^2 2^2的前n 项之和编写程序

裂项an=(n+2)/[n!+(n+1)!+(n+2)!]=（n+2)[n!(1+n+1+(n+1)(n+2))]=（n+2)/[n!(n+2)^2]=1/[n!(n+2)]=(n+1)/(n+2)!

a=(2n-1)×2＾(n-1)是这个吗?Sn=1×1+3×2+5×4+……+(2n-1)×2＾(n-1)2Sn=1×2+3×4+5×8+……+（2n-3）×2＾（n-1）+(2n-1)×2＾n相减2

an=1/[(2n+1)(2n+3)]=[(2n+3)-(2n+1)]/[2(2n+1)(2n+3)]=(2n+3)/[2(2n+1)(2n+3)]-(2n+1)/[2(2n+1)(2n+3)]=1/

an=(n-1)*2^(n-1)sn=(1-1)*2^(1-1)+(2-1)*2^(2-1)+.+(n-1)*2^(n-1)2sn=2*(1-1)*2^(1-1)+2*(2-1)*2^(2-1)+.+

an=(2n-1)(1/4)^n=n(1/4)^(n-1)-(1/4)^nSn=a1+a2+..+an=[summation(i:1->n){i(1/4)^(i-1)}]-(1/3)(1-(1/4)^

老式写法longint格式用%ld--l是L小写.现在写%d就可以了,longint,shortint,int都用%d程序用ASCII码写成：#includemain(){longs,n,k,i;sc

M=1＋2＋3＋…＋n=[n(n＋1)]/2N=1²＋2²＋3²＋…＋n²=[n(n＋1)(2n＋1)]/6P=1³＋2³＋3³＋

Tn=b1+b2+…+bn=[k+k^3+k^5+…+K^(2n-1)]+2(1+2+…+n)=k[k^(2n)-1]/(k^2-1)+n(n+1)

sn=3*3^1+5*3^2+.+(2n+1)*3^n①3sn=3*3^2+5*3^3+.+(2n-1)*3^n+(2n+1)*3^(n+1)②①-②-2Sn=Sn-3Sn=-2n*3^(n+1),因

#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]

an=Sn-Sn-1=1/3n(n+1)(n+2)-1/3n(n+1)(n-1)=n(n+1)所以1/an=1/n(n+1)=1/n-1/n+1数列(1/an)的前n项和=1-1/2+1/2-1/3+

sn=1/2+2/4+3/8...n/2^nsn/2=1/4+2/8...+n/2^(n+1)两式相减,得sn/2=1/2+1/4+1/8...+1/2^n-n/2^(n+1)=1-1/2^n-n/2

当n=1时,a1+a1=1/2(1*1+5*1+2)=4a1=2当n=2时a1+a2+a2=1/2(2*2+5*2+2)2+2*a2=8a2=3当n=3时,a1+a2+a3+a3=1/2(3*3+5*

S2=a1+a2=1+a2=2²×a23a2=1a2=1/3S3=a1+a2+a3=1+1/3+a3=3²×a38a3=4/3a3=1/6a1=1=2/[1×(1+1)]a2=1/

S=0.25n(n+1)(n+2)(n+3)再问：能提供方法么？谢谢！是用裂项么？再答：n(n+1)(n+2)=0.25[n(n+1)(n+2)(n+3)-(n-1)n(n+1)(n+2)]

因为(n+1)^3-n^3=(n+1-n)[(n+1)^2+n(n+1)+n^2]=3n^2+3n+1所以3n^2=(n+1)^3-n^3-3n-1所以3*1^2+3*2^2+……+3n^2=[(1+

错位相减Sn=n*2^(n+1)

第一个问题答案1（n=1)n2\(n-1)2(n大于等于2）第二个问题答案Sn+1－3Sn+2Sn－1=0(n∈N*)等价于(Sn+1-Sn)-2(Sn-Sn-1)=0(n∈N*)等价于(an+1)-

底数是n+1吗?Logn+1(n+2)=Lg(n+2)/Lg(n+1)a1a2…an=(lg3/lg2)*(lg4/lg3)*(lg5/lg4)…[lg(n+1)/lgn]*[lg(n+2)/lg(n

这是典型的错位相减求和,要举一反三!你拿张纸,先把Sn求和表达式写出来,要求写出a1+a2…+an-1+an四个就行;接着再起一行,写出2Sn的表达式,也写出2a1+2a2…+2an-1+2an就行.