数列an满足a1等于1,nan 1
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an/n-a(n+1)/n+1=2/n(n+1)=2(1/n-1/n+1)………………a1/1-a2/2=2(1/1-1/2)a1-an/n=2(1/1-1/2+1/2-1/3+……+1/(n-1)-
以下用a(n)表示数列的第n项.题目中的式子是a(n)=a(n-1)/{[(-1)^n]×a(n-1)-2}的意思吧?(1)由a(n)=a(n-1)/[(-1)^n×a(n-1)-2],两边取倒数,得
a1+2a2+3a3+...+(n-1)a(n-1)=(n-1)n(n+1)a1+2a2+3a3+...+nan=n(n+1)(n+2)2试-1式得nan=3n(n+1)an=3(n+1)
(1)当n=1时,a1>=3=1+2,an>=n+2成立;当n>1时,an=(an-1)^2-nan-1+1,令S=an-(n+2)=(an-1)^2-nan-1+1-(n+2)=(an-1)^2-(
a1+2a2+3a3+...+nan=n(n+1)*(n+2),则:a1+2a2+3a3+...+(n-1)×an-1=n(n-1)*(n+1),两式相减:nan=n(n+1)*(n+2)-n(n-1
令n=1时,a1=1*2*3=6;依题意:a1+2a2+3a3+.+nan=n(n+1)(n+2),a1+2a2+3a3+.+nan+(n+1)a(n+1)=(n+1)(n+2)(n+3)两式相减,得
∵2nan+1=(n+1)an,∴a(n+1)/an=(n+1)/2n,∴a2/a1=2/2a3/a2=3/2×2a4/a3=4/2×3a5/a4=5/2×4……an/a(n-1)=n/2(n-1)两
∵数列{a[n]}满足a[1]+2a[2]+3a[3]+...+na[n]=(n+1)(n+2)∴a[1]+2a[2]+3a[3]+...+na[n]+(n+1)a[n+1]=(n+2)(n+3)将上
解An+1/An=2^n所以A2/A1=2所以数列是以1为首相2为公比的等比数列所以通向公式an=2^(n-1)
na(n+1)=(n+1)ana(n+1)/an=(n+1)/n1由1式可以推出an/a(n-1)=n/(n-1).a2/a1=2/1左边相乘,右边相乘,相互约分得a(n+1)/a1=(n+1)/1a
n=1时,a1=3;n>1时,a1+2a2+3a3+...+(n-1)a(n-1)=(2n-3)*3^(n-1)an=【(2n-1)*3^n-(2n-3)*3^(n-1)】/n自己化简吧
na(n+1)=(n+1)ana(n+1)/an=(n+1)/n1由1式可以推出an/a(n-1)=n/(n-1).a2/a1=2/1左边相乘,右边相乘,相互约分得a(n+1)/a1=(n+1)/1a
a1+2a2+3a3+~+nan=n(n+1)(2n+1)知,a1+2a2+3a3+~+(n-1)an-1=(n-1)n(2n-1),n≠1时两式相减知an=(n+1)(2n+1)-(n-1)(2n-
a1+2a2+3a3+...+nan=n(n+1)(2n+1)①a1+2a2+3a3+...+(n-1)a(n-1)=n(n-1)^2②①-②可得:nan=n(n+1)(2n+1)-n(n-1)^2=
已知数列{an}满足a1=1,a(n+1)=nan(1)求{an}的通项公式;(2)证明:1/a1+1/a2+.+1/an≤3-(1/2)^(n-2).(1)因为a(n+1)=nan,即a(n+1)/
(1)证明:由bn=3-nan得an=3nbn,则an+1=3n+1bn+1.代入an+1-3an=3n中,得3n+1bn+1-3n+1bn=3n,即得bn+1-bn=13.所以数列{bn}是等差数列
(1)用数学归纳法.A(n+1)=An^2-nAn+1=An(An-n)+1>=An*2+1>=(n+2)*2+1=2n+5>n+1+2(2)因为an>=n+2,所以an-n>=2A(n+1)=An(
证明:因为:a1+2a2+3a3+…+nan=n(n+1)(n+2),记:bn=nan,那么:b1+b2+...+bn=n(n+1)(n+2)将n-1带入,得:b1+b2+...+b(n-1)=(n-
A2=A1+1A3=A2+2A4=A3+3.An=A(n-1)+(N-1)左式上下相加=右式上下相加An=A1+[1+2+3+...+(N-1)]An=1+[N(N-1)]/2