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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+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

(1) a(n+1)=an/(an+1) 1/a(n+1) = (an+1)/an 1/a(n+1) -1/an = 1 =>(1/an)是等差数列 1/an -1/a1= n-1 1/an =n an =1/n (2) bn =1/(2^n.an) = (1/2)[n(1/2)^(n-1)] consider 1+x+x^2+..+x^n = (x^(n+1)- 1)/(x-1) 1+2x+..+nx^(n-1) =[(x^(n+1)- ...

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...

(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...

解本题其实是很有技巧性的,而并不是善解人意一说的什么没技巧可言。运用拆项法很容易解决。 an=n(n+1)²=n(n+1)(n+2-1)=n(n+1)(n+2)-n(n+1) Sn=a1+a2+...+an =[1×2×3+2×3×4+...+n(n+1)(n+2)]-[1×2+2×3+...+n(n+1)] =¼×[1×2×3×4-0×1×2×...

裂项相减 1/[n(n+1)]=1/n-1/(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

a(n+1)/an=n+2/n改成a(n+1)/an=(n+2)/n 这是a(n+1)/an=f(n)的形式, 用累乘法: a2/a1=3/1 a3/a2=4/2 ... a(n-1)/a(n-2)=n/(n-2) an/a(n-1)=(n+1)/(n-1) 等式左边相乘=等式右边相乘得 an/a1=n(n+1)/2 an=n(n+1)/2

∵a(n+1)-an=an/(n+1) ∴a(n+1)=an+an/(n+1) =an*(n+2)/(n+1) ∴a(n+1)/an=(n+2)/(n+1) 那么an/a(n-1)=(n+1)/n a(n-1)/a(n-2)=n/(n-1) ………………………… a3/a2=4/3 a2/a1=3/2 累乘,得:an/a1=(n+1)/2 而a1=1,∴an=(n+1)/2

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