Math topic中文

Relationship un sn

Relationship between unu_n (the nnth term) and SnS_n (the sum to nn terms).

Practice questions

  1. 1The sum of the first nn terms of a series is given by Sn=3n(n+2)S_n = 3n(n + 2). Show that this series follows an arithmetic pr…
  2. 2The sum of the first nn terms of a series is given by Sn=3n(n+2)S_n = 3n(n + 2). Show that this series follows an arithmetic pr…
  3. 3Consider Sn=21×2×3+22×3×4+23×4×5+…+2n(n+1)(n+2)S_{n}=\dfrac{2}{1\times2\times3}+\dfrac{2}{2\times3\times4}+\dfrac{2}{3\times4\times5}+\ldots+\dfrac{2}{n(n+1)(n+2)}…
  4. 4已知{an}\left\{{a}_{n}\right\}是公差为22的等差数列,Sn{S}_{n}为{an}\left\{{a}_{n}\right\}的前nn项和,若S5=S3S_{5}=S_{3},则a4=()a_{4}=(\quad)…
  5. 5A sequence of numbers u1,u2,u3,…u_{1}, u_{2}, u_{3}, \ldots has sum SnS_{n} where Sn=∑r=1nurS_{n} = \sum_{r=1}^{n} u_{r}. It is given th…
  6. 6A sequence of numbers u1,u2,u3,…u_1, u_2, u_3, \dots has a sum SnS_n where Sn=∑r=1nurS_n = \sum_{r=1}^{n} u_r. It is given that Sn=An2(n+1)−BnS_n = An^2(n + 1) - Bn…
  7. 7等差数列{an}\{a_{n}\}的前nn项和为SnS_{n},数列{bn}\{b_{n}\}为等比数列,Sn=n2+nS_{n}\text{=}n^{2}\text{+}n,b1=a1,b2=a1a22.b_{1}\text{=}a_{1}\text{,}b_{2}\text{=}\dfrac{a_{1}a_{2}}{2}\text{.}…
  8. 8The sum, SnS_n, of the first nn terms of the sequence u1,u2,u3,…u_1, u_2, u_3, \dots is given by Sn=n2+4n.S_n = n^2 + 4n . Find unu_n…
  9. 9已知数列{an}\{a_{n}\}的前nn项和为SnS_{n},首项为a1a_{1},且44,ana_{n},SnS_{n}成等差数列. (1)求数列{an}\{a_{n}\}的通项公式; (2)若an2=2bna_{n}^{2}={{2}^{{{b}_{n}}}}…
  10. 10数列{an}\{a_{n}\}中,Sn=4−an−12n−2S_{n}=4-a_{n}- \dfrac {1}{2^{n-2}}. ((Ⅰ))求a1a_{1},a2a_{2},a3a_{3},a4a_{4}; ((Ⅱ))…
  11. 11已知首项为11的数列{an}\{a_{n}\}的前nn项和为SnS_{n},且an+1=13Sna_{n\text{+}1}\text{=}\dfrac{1}{3}S_{n},数列{bn}\{b_{n}\}满足bn=log⁡3anb_{n}\text{=}\log_{3}a_{n}…
  12. 12A sequence u1,u2,u3,…u_1, u_2, u_3, \dots is such that u1=12u_1 = \frac{1}{2} and un+1=un−n2+n−1(n+2)!,for all n≥1.u_{n+1} = u_n - \frac{n^2 + n - 1}{(n+2)!}, \quad \text{for all } n \ge 1.…
  13. 13已知在数列{an}\{a_{n}\}中,a1=1a_{1}=1,其前nn项和为sns_{n},且an=2sn22sn−1(n⩾2)(1)a_{n}= \dfrac {2 s_{ n }^{ 2 }}{2s_{n}-1}(n\geqslant 2)(1)证明{1sn}\{ \dfrac {1}{s_{n}}\}…
  14. 14已知SnS_{n}为数列{an}\{a_{n}\}的前nn项和,Sn=1−anS_{n}=1-a_{n},则S5=()S_{5}=(\quad) A. 3116\dfrac{31}{16} B. 312\dfrac{31}{2} C.…
  15. 15已知公差不为00的等差数列{an}\{a_{n}\}的前nn项和为SnS_{n},且2Sn=an+1an2S_{n}=a_{n+1}a_{n},则a1=a_{1}= ___ .
  16. 16已知数列{an}\{a_{n}\}的前nn项为和SnS_{n},点(n,Snn)(n, \dfrac {S_{n}}{n})在直线y=12x+112y= \dfrac {1}{2}x+ \dfrac {11}{2}上.数列{bn}\{b_{n}\}…
  17. 17已知各项均为正数的数列{an}\{a_{n}\}的前nn项和为SnS_{n},且a1=1a_{1}=1,an+1(an+1−1)=an(an+1).a_{n+1}(a_{n+1}-1)=a_{n}(a_{n}+1).若[x][x]表示不超过xx的最大整数,bn=[(n+1)22Sn]b_{n}=[\dfrac{(n+1)^{2}}{2S_{n}}]…
  18. 18已知数列{an}\{a_{n}\}中,a1=1a_{1}=1,a2=2a_{2}=2,且an+1=4an−3an−1(n∈N∗,n⩾2)a_{n+1}=4a_{n}-3a_{n-1}(n∈N^{*},n\geqslant 2). ((Ⅰ))令bn=an+1−anb_{n}=a_{n+1}-a_{n}…
  19. 19A sequence u1,u2,u3,…u_1, u_2, u_3, \dots is such that u1=e10u_1 = \dfrac{e}{10} and un−un−1=e2n−1u_n - u_{n-1} = \dfrac{e}{2^{n-1}}, for all n≥2n \ge 2…
  20. 20平面内有nn条直线,最多可将平面分成f(n)f(n)个区域,则f(n)f(n)的表达式为()(\quad) A. n+1          n+1\:\:\:\:\:\:\:\:\:\: B. 2n          2n\:\:\:\:\:\:\:\:\:\: …
  21. 21已知数列{an}\{a_{n}\}中,a1=2a_{1}=2,a2=3a_{2}=3,其前nn项和SnS_{n}满足Sn+1+Sn−1=2Sn+1S_{n+1}+S_{n-1}=2S_{n}+1,其中n⩾2n\geqslant 2,n∈N∗n∈N^{*}…
  22. 22The sum, SnS_{n}, of the first nn terms of a sequence u1,u2,u3,…u_{1}, u_{2}, u_{3}, \dots is given by Sn=a(3n)+bn+cS_{n}=a(3^{n})+bn+c, w…
  23. 23已知Sn{S}_{n}为数列{an}\left\{{a}_{n}\right\}的前nn项和,a1=1{a}_{1}=1,2Sn=(n+1)an2{S}_{n}=(n+1){a}_{n},则a5={a}_{5}=___;若存在唯一的正整数nn…
  24. 24The ninth, fifth and second terms of an arithmetic progression are successive terms of a geometric progression with firs…
  25. 25Ryan has playing cards which he stacks into pyramids. He will begin by stacking up cards to form a pyramid with 1 level,…
  26. 26设数列{an}\{a_{n}\}的前nn项和为SnS_{n},且 Sn=n2−4n+4S_{n}=n^{2}-4n+4. (1)求数列{an}\{a_{n}\}的通项公式; (2)设bn=an2nb_{n}= \dfrac {a_{n}}{2^{n}}…
  27. 27下列结论:正确的序号是 ___ . ①△ABC①\triangle ABC中,若A>BA > B则一定有sin⁡A>sin⁡B\sin A > \sin B成立; ②②数列{an}\{a_{n}\}的前nn项和Sn=n2−2n+1S_{n}=n^{2}-2n+1…
  28. 28A sequence is such that u0=2u_0 = 2 and un=un−1+n3+(12)nu_n = u_{n-1} + n^3 + \left(\frac{1}{2}\right)^n for n≥1n \ge 1. It is given that…
  29. 29已知数列{an}\{a_{n}\}的前nn项和为SnS_{n},Sn=2an−3n(n∈N∗).(1)S_{n}=2a_{n}-3n(n∈N^{*}).(1)证明数列{an+3}\{a_{n}+3\}是等比数列,求出数列{an}\{a_{n}\}的通项公式; (2…
  30. 30数列{an}\{a_{n}\}的前nn项和是SnS_{n},a1=1a_{1}=1,2Sn=an+1(n∈N+)2S_{n}=a_{n+1}(n∈N_{+}),则an=a_{n}= ___.
  31. 31The ninth, fifth and second terms of an arithmetic progression are successive terms of a geometric progression with firs…
  32. 32The sum of the first nn terms of a sequence, uru_r, is given by ∑r=1nur=1−n(n+1)!\displaystyle\sum_{r=1}^n u_r = 1 - \dfrac{n}{(n+1)!}…
  33. 33The sum, SnS_n, of the first nn terms of a geometric progression is given by Sn=6−23n−1S_n = 6 - \dfrac{2}{3^{n-1}}. Find the f…
  34. 34已知向量a→=(2,−n) \overrightarrow{a}=(2,-n),b→=(Sn,n+1) \overrightarrow{b}=(S_{n},n+1),n∈N∗n∈N^{*},其中SnS_{n}是数列{an}\{a_{n}\}的前nn项和,…