an=(2n-1) (1 2)的n次方求sn分组求和
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/18 22:16:07
1.an-an-1=2(n-1)-1=2(n-1)2n-2=-12n=2-12n=1n=1/22.3+(n-1)(-2)=-2n-53-2n+2=-2n-55=-5题目有错,无解.3.2+(n-1)x
an=n/(n-1)×a(n-1)+2n×3^(n-2)∴an/n=a(n-1)/(n-1)+2×3^(n-2)------(1)a(n-1)/(n-1)=a(n-2)/(n-2)+2×3^(n-3)
M=1+2+3+…+n=[n(n+1)]/2N=1²+2²+3²+…+n²=[n(n+1)(2n+1)]/6P=1³+2³+3³+
(1)由已知a2=2a1+2,a3=2a2+3=4a1+7,若{an}是等差数列,则2a2=a1+a3,即4a1+4=5a1+7,得a1=-3,a2=-4,故d=-1. &nbs
An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn
求数列{an}前n项的和,常用的方法就是裂项相消法.因为an=n(n+1)=n(n+1)[(n+2)-(n-1)]/3=[n(n+1)(n+2)-(n-1)n(n+1)]/3=(1/3)[-(n-1)
An=1/(n+1)+1/(n+2)+…+1/(2n-1)+1/(2n)则An+1=1/(n+2)+1/(n+3)+…+1/(2n-1)+1/(2n)+1/(2n+1)+1/(2n+2)则An+1-A
(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
a1=33,a(n)-a(n-1)=2(n-1),a(n)=a1+(a2-a1)+(a3-a2)+……+(a(n)-a(n-1))=33+2+2×2+……+2(n-1)=33+n(n-1).an/n=
C(k,n)ak=n!/((n-k)!*k!)*(k(k+1))/2=(n-1)!/((n-k)!(k-1)!)*(n(k+1))/2=C(k-1,n-1)*n/2*(k+1)An=n/2*[C(0,
an=1/n(n+1)(n+2)=[1/n(n+1)-1/(n+1)(n+2)]/2,a1=1/6所以S1=a1=1/6n>=2时,Sn=a1+a2+...+an=[1/1*2-1/2*3]/2+[1
1.如果An=n+(1/3)^nSn=n(n+1)/2+(1/3)×(1-1/3^n)/(1-1/3)=n(n+1)/2+(1-1/3^n)/2如果An=(n+1)/3^nSn=A1+A2+A3+……
(1)证明:∵在数列{a[n]}中,已知a[n]+a[n+1]=2n(n∈N*)∴用待定系数法,有:a[n+1]+x(n+1)+y=-(a[n]+xn+y)∵-2x=2,-x-2y=0∴x=-1,y=
【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+……
2Sn=(n+1)an2S(n-1)=na(n-1)两式相减得2an=(n+1)an-na(n-1)移相得(1-n)an=-na(n-1)得an=(n/(n-1))a(n-1)an=(n/(n-1))
/>n≥2时,an=Sn/n+2(n-1)Sn=nan-2n(n-1)S(n-1)=(n-1)an-2(n-1)(n-2)Sn-S(n-1)=an=nan-2n(n-1)-(n-1)an+2(n-1)
a(n+1)-an=2n是一个递推关系式,同理,有an-a(n-1)=2(n-1),...以此类推,把这些式子依次相加,左后一个式子为a2-a1=2,所以,前后项都可以抵消一部分,你自己列一下就知道了
an=1/(√(n+2)+√n)=[√(n+2)-√n]/[(√(n+2)+√n)(√(n+2)-√n)]=[√(n+2)-√n]/(n+2)-n)=[√(n+2)-√n]/22an=√(n+2)-√
an=(3n-2)/(3n+1)a10=(3*10-2)/(3*10+1)=28/31(3n-2)/(3n+1)=7/107(3n+1)=10(3n-2)21n+7=30n-2030n-21n=7+2