数列1 n(n 1)的前2017项的和为

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

解题思路：利用等差数列的性质求解。解题过程：最终答案：略

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)^

an=2^n+n+1Sn=a1+a2+...+an=(2^1+1+1)+(2^2+2+1)+...+(2^n+n+1)=(2^1+2^2...+2^n)+(1+2+...+n)+(1+1+...+1)

an=[(n+1)^2+1]/[(n+1)^2-1]=1+2/[(n+1)^2-1]=1+2/[n(n+2)]=1+[1/n-1/(n+2)]Sn=a1+a2+...+an=n+[1+1/2-1/(n

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

典型的“等差*等比”型数列,用错位相减法Sn=2*3^1+3*3^2+4*3^3+...+(n+1)*3^n(1)3Sn=2*3^2+3*3^3+...+n*3^n+(n+1)*[3^(n+1)](2

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

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+

(1)令n=1a1=S1=32-1+1=32Sn=32n-n²+1Sn-1=32(n-1)-(n-1)²+1an=Sn-Sn-1=32n-n²+1-32(n-1)+(n-

Sn=a1+a2+~+an=1+1/3^1+1+2+1/3^2+1+~+n+1/3^n+1=n(1+n)/2+[1/3×(1-1/3^n)]/(1-1/3)+n=(n^2-1/3^n+3n+1)/2

n=n(n+1)=n^2+nSn=b1+b2+...+bn=(1^2+1)+(2^2+2)+...+(n^2+n)=(1^2+2^2+...+n^2)+(1+2+...+n)=n(n+1)(2n+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)]

错位相减Sn=n*2^(n+1)

当n≥2时an=Sn-S(n-1)=(3n²+n+1)-[3(n-1)²+(n-1)+1]=3n²+n+1-[3(n²-2n+1)+n-1+1]=3n²

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.