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n+1 An+n

解: a(n+1)-an=n+1=½[(n+1)²-n²]+½ [a(n+1)-½(n+1)²]-(an-½n²)=½,为定值 a1-½×1²=1-½=½ 数列{an-½n²}是以½为首项,½为公差的等差数列 an-½n...

A(n+1)-An=n+1 An-A(n-1)=n A(n-1)-A(n-2)=n-1 ............. A2-A1=2 各式相加得:A(n+1)-A1=2+3+4+...+n+(n+1) =(2+n+1)n/2 =(n+3)n/2 故 An=(n+2)(n-1)/2+2

a(n+1)=an+1/n-1/(n+1) a(n+1)+1/(n+1)=an+1/n 设bn=an+1/n 则:b(n+1)=bn b1=a1+1/1=2 所以:数列bn为常数列bn=2 所以an+1/n=2 an=2-1/n

a[n+1]-a[n] > 1/4 1/(1-a[n]) - a[n] = (1-2a[n])^2/(4-4a[n]) > 0 a[n]单增 又a[n]=1/4 (极限时通常严格不等式要变成非严格不等式) (A-1/2)^2

解答:证明(Ⅰ)∵nan+1=(n+1)an+n(n+1),∴an+1n+1=ann+1,∴an+1n+1?ann=1,∴数列{ann}是以1为首项,以1为公差的等差数列;(Ⅱ)由(Ⅰ)知,ann=1+(n?1)?1=n,∴an=n2,bn=3n?an=n?3n,∴Sn=1×3+2×32+3×33+…+(n?1)?3n-1+n?3n①3Sn=1×32+2...

a(n+1)=(n+1)a(n)/n + (n+1)/2^n, a(n+1)/(n+1)=a(n)/n + 1/2^n, b(n+1)=b(n) + 1/2^n, b(1)=a(1)/1=1. 2^nb(n+1)=2*[2^(n-1)b(n)] + 1, 2^nb(n+1) + 1 = 2[2^(n-1)b(n) + 1], {2^(n-1)b(n)+1}是首项为b(1)+1=2,公比为2的等比数列. 2^(n-1)b(n)+...

∵Sn=1/2(Sn-S(n-1)+n/(Sn-S(n-1)) ∴Sn^2-S(n-1)^2=n 数列{Sn^2-S(n-1)^2}是自然数数列 其前n项和=(S2^2-S1^2)+(S3^2-S2^2)+...+(Sn^2-S(n-1))=Sn^2-S1^2=2+3+...+n=n(n+1)/2-1 ∴Sn^2=n(n+1)/2 Sn=√(n(n+1)/2) (an>0=>Sn>0) an=Sn-S(n-1)=√(n(n+1...

a(n+1)/an=n/n+1 则an/a(n-1)=(n-1)/n a(n-1)/a(n-2)=(n-2)/(n-1) .......... a2/a1=1/2 叠乘 an/a1=1/n 因a1=1 故an=1/n

(n+1)a(n+1)=an+n a(n+1)=(an+n)/(n+1) a(n+1)-1=(an+n-n-1)/(n+1)=(an-1)/(n+1) [a(n+1)-1]/(an-1)=1/(n+1) (an-1)/[a(n-1)-1]=1/n [a(n-1)-1]/[a(n-2)-1]=1/(n-1) ………… (a2-1)/(a1-1)=1/2 连乘 (an-1)/(a1-1)=(1/2)(1/3)...(1/n)=1/n! an-1=(a...

an/a(n-1)=[n-1]/[n+1] a2/a1=1/3 a3/a2=2/4 a4/a3=3/5 . . . a(n-1)/a(n-2)=n-2/n an/a(n-1)=n-1/n+1 相乘:an/a1=2/n*(n+1) an=2/n*(n+1)=2*[1/n-1/(n+1)] Sn=2[1-1/2+1/2-1/3+...+1/n-1/(n+1)]=2[1-1/(n+1)]=2n/(n+1)

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