数列[1+2^n-1,]的前N项和为答案
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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)^
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)
M=1+2+3+…+n=[n(n+1)]/2N=1²+2²+3²+…+n²=[n(n+1)(2n+1)]/6P=1³+2³+3³+
(1/3)n(3n+2)=(1/3)n(3n+3)-(1/3)n=n(n+1)-n/3=(1/3)[n(n+1)(n+2)-(n-1)n(n+1)]-(1/6)[n(n+1)-(n-1)n](1/3)
求a‹n›=(3n+1)(2^n/3)的前n项和S‹n›=(1/3)[(4×2)+(7×2²)+(10×2³)+(13×2̾
(1)当n≥2时,an=Sn-Sn-1=n(2n-1)-(n-1)(2n-3)=4n-3,当n=1时,a1=S1=1,适合.∴an=4n-3,∵an-an-1=4(n≥2),∴an为等差数列.(2)由
用错位相减法:sn=1*3^1+3*3^2+5*3^3+.+(2n-1)*3^n3*sn=1*3^2+3*3^3+.+(2n-3)*3^n+(2n-1)*3^(n+1)-2sn=1*3^1+2*3^2
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
sn=3*1-1+2^1+3*2-1+2^2+.+3n-1+2^n=3*(1+2+.+n)-n+2^1+2^2+.+2^n=3n(n+1)/2-n+2*(1-2^n)/(1-2)=(3n^2+3n-2
【方法1:强行展开a(n)表达式】1+2+……+n=n(n+1)/21^2+2^2+……+n^2=n(n+1)(2n+1)/61^3+2^3+……+n^3=n^2(n+1)^2/41^4+2^4+……
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)
因为an=(2n-1)/2^n=n/2^(n-1)-1/2^n设数列{n/2^(n-1)}前n项和为Tn,数列{1/2^n}前n项和为Pn,则Sn=Tn-PnTn=1+2/2+3/2²+4/
Tn=1*1+2*2^1+3*2^2+……+(n-1)*2^(n-2)+n*2(n-1)2Tn=2*1+2*2^2+3*2^3+……+(n-1)*2(n-1)+n*2^n下减上Tn=-1-2-2^2-
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.