Math topic中文

Sum difference of series

Sum and difference of two series.

Practice questions

  1. 1Given f(r)=r(r−1)!f(r)=\dfrac{r}{(r-1)!}, show that f(r−1)−f(r)=r2−3r+1(r−1)!f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}. Hence find ∑r=2nr2−3r+1(r−1)!.\displaystyle\sum_{r=2}^{n} \frac{r^2-3r+1}{(r-1)!}.…
  2. 2已知数列{an}\{a_{n}\}中，a1=76a_{1}=\dfrac{7}{6}，an2−an+1=an+1(n∈N∗)a_{n}^{2}-a_{n}+1=a_{n+1}(n\in N^{*})，SnS_{n}是{1an}\{\dfrac{1}{a_{n}}\}的前…
  3. 3Given f(r)=r(r−1)!f(r)=\dfrac{r}{(r-1)!}, show that f(r−1)−f(r)=r2−3r+1(r−1)!f(r-1)-f(r)=\dfrac{r^2-3r+1}{(r-1)!}. Hence find ∑r=2nr2−3r+1(r−1)!\displaystyle\sum_{r=2}^n \dfrac{r^2-3r+1}{(r-1)!}…
  4. 4Show that 32r+1−42r+3+12r+5=8r+k(2r+1)(2r+3)(2r+5)\frac{3}{2r+1}-\frac{4}{2r+3}+\frac{1}{2r+5}=\frac{8r+k}{(2r+1)(2r+3)(2r+5)}, where kk is a constant to be…
  5. 5Prove by induction that, for all positive integers nn and for x∈Rx\in\mathbb{R}, x≠0, x≠1x\ne 0,\ x\ne 1, 1x−1−1x−1x2−⋯−1xn=1xn(x−1) .\frac{1}{x-1}-\frac{1}{x}-\frac{1}{x^2}-\dots-\frac{1}{x^n}=\frac{1}{x^n(x-1)}\,.…
  6. 6Using the formulae for sin⁡(A±B)\sin(A \pm B), prove that sin⁡(2r+1)θ−sin⁡(2r−1)θ=2cos⁡2rθsin⁡θ.\sin(2r + 1)\theta - \sin(2r - 1)\theta = 2 \cos 2r\theta \sin \theta .…
  7. 7By writing 3(3r+1)(3r+4)\frac{3}{(3r+1)(3r+4)} in partial fractions, find ∑r=1n3(3r+1)(3r+4)\sum_{r=1}^{n} \frac{3}{(3r+1)(3r+4)} in terms of nn.…
  8. 8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{.}…
  9. 9Verify that 3(r+1)!−2r!−1(r−1)!=−r2+3r+1(r+1)!\dfrac{3}{(r+1)!} - \dfrac{2}{r!} - \dfrac{1}{(r-1)!} = -\dfrac{r^2 + 3r + 1}{(r+1)!}. Hence find ∑r=1n−r2−3r+1(r+1)!.\displaystyle\sum_{r=1}^{n} \dfrac{-r^2 - 3r + 1}{(r+1)!}. …
  10. 10有一列按一定顺序和规律排列的数： 第11个数是11×2\dfrac{1}{1×2}；第22个数是12×3\dfrac{1}{2×3}；第33个数是13×4… \dfrac{1}{3×4} …对任何正整数nn，第…
  11. 11Show that 7n−2−5n−1−2n=9n−4(n−2)(n−1)n\frac{7}{n-2} - \frac{5}{n-1} - \frac{2}{n} = \frac{9n-4}{(n-2)(n-1)n}. Hence find ∑n=3N9n−4(n−2)(n−1)n\sum_{n=3}^{N} \frac{9n-4}{(n-2)(n-1)n}…
  12. 12A sequence is such that u0=2u_{0}=2 and un=un−1+n3+(12)nu_{n}=u_{n-1}+n^{3}+\left(\dfrac{1}{2}\right)^{n} for n≥1n\ge 1. (a) It is given…
  13. 13A 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.…
  14. 14It is given that f(r)=(r−2)2r−1f(r)=(r-2)2^{r-1}. Show that f(r+1)−f(r)=2rarf(r+1)-f(r)=2^r a r, where aa is a constant to be determined. Hence f…
  15. 15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}}\}…
  16. 16By writing 2(2r+1)(2r+3)\frac{2}{(2r+1)(2r+3)} in partial fractions, find an expression for ∑r=2n2(2r+1)(2r+3)\sum_{r=2}^{n} \frac{2}{(2r+1)(2r+3)}…
  17. 17By writing 2r(r2−1)\dfrac{2}{r(r^2-1)} in partial fractions, find an expression for ∑r=2n2r(r2−1)\displaystyle\sum_{r=2}^{n}\dfrac{2}{r(r^2-1)}…
  18. 18Write 44r2+12r+5\dfrac{4}{4r^{2}+12r+5} in partial fractions. Find ∑r=2n44r2+12r+5\displaystyle\sum_{r=2}^{n} \dfrac{4}{4r^{2}+12r+5}. A sequ…
  19. 19阅读下面问题： 12+1=1×(2−1)(2+1)(2−1)=2−1\dfrac{1}{ \sqrt{2}+1} = \dfrac{1×\left( \sqrt{2}-1\right)}{\left( \sqrt{2}+1\right)\left( \sqrt{2}-1\right)} = \sqrt{2}-1…
  20. 20A sequence u1,u2,u3,…u_1, u_2, u_3, \dots is such that u1=0u_1 = 0 and un+1=un+4+n−n2[(n+1)(n+2)]2, for all n∈Z+.u_{n+1} = u_n + \frac{4 + n - n^2}{[(n+1)(n+2)]^2}, \text{ for all } n \in \mathbb{Z}^+.…
  21. 21Find how many positive integers, less than 1000, are odd numbers. odd numbers which are not divisible by 5. Find the sum…
  22. 22Given that ∑n=1Nn3=[N2(N+1)]2,\sum_{n=1}^{N} n^{3} = \left[\frac{N}{2}(N+1)\right]^{2}, Find 12−23+33−43+⋯−(2N)3,1^{2}-2^{3}+3^{3}-4^{3}+\dots-(2N)^{3},…
  23. 23Prove that 2n+1n2+2n+n2−1=n2+2n−n2−1.\displaystyle \frac{2n+1}{\sqrt{n^2+2n+\sqrt{n^2-1}}} = \sqrt{n^2+2n} - \sqrt{n^2-1}. Hence find ∑n=1N2n+1n2+2n+n2−1\displaystyle \sum_{n=1}^N \frac{2n+1}{\sqrt{n^2+2n}+\sqrt{n^2-1}}…
  24. 24Given that 1−2r=A(r+1)+Br1-2r = A(r+1) + Br, find the constants AA and BB. Use the method of differences to find ∑r=1n1−2r3r\displaystyle \sum_{r=1}^n \frac{1-2r}{3^r}…
  25. 25Verify that 3(r+1)!−2r!−1(r−1)!=−r2−3r+1(r+1)!\displaystyle \frac{3}{(r+1)!}-\frac{2}{r!}-\frac{1}{(r-1)!}=\displaystyle\frac{-r^2-3r+1}{(r+1)!}. Hence…
  26. 26Find the exact value of π∫16x3 dx\pi \int_{1}^{6} x^3 \text{ d}x. The diagram shows a sketch of the curve y=x32y = x^{\frac{3}{2}}…
  27. 27Prove by induction that ∑r=1nr3=14n2(n+1)2.\sum_{r=1}^{n} r^{3} = \frac{1}{4} n^{2} (n+1)^{2}. The r th term of a series is given by ln⁡(2ar)\ln\left(2a^{r}\right)…
  28. 28The reproduction number, R0R_0, of a disease is a mathematical term that indicates how contagious an infectious disease…
  29. 29Use the standard series for ln⁡(1+x)\ln(1+x) and ln⁡(1−x)\ln(1-x) to find the first three non-zero terms in the series for ln⁡(1+x1−x).\displaystyle\ln\left(\frac{1+x}{1-x}\right).…
  30. 30阅读并解答后面的问题． 11×2=12 \dfrac {1}{1×2}= \dfrac {1}{2}，11−12=12 \dfrac {1}{1}- \dfrac {1}{2}= \dfrac {1}{2}；12×3=16 \dfrac {1}{2×3}= \dfrac {1}{6}…
  31. 31Given that z1=r1(cos⁡θ1+isin⁡θ1)z_1 = r_1(\cos\theta_1 + \mathrm{i}\sin\theta_1) and z2=r2(cos⁡θ2+isin⁡θ2)z_2 = r_2(\cos\theta_2 + \mathrm{i}\sin\theta_2), pr…
  32. 32Show that 1a+b=a−ba−b\dfrac{1}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}-\sqrt{b}}{a-b}. Find ∑r=1250014r−3+4r+1\displaystyle\sum_{r=1}^{2500}\dfrac{1}{\sqrt{4r-3}+\sqrt{4r+1}}…
  33. 33The graph of y=x+40y = \sqrt{x+40} is shown in the diagram below. By considering the shaded rectangle, show that n+40<∫nn+1x+40 dx.\sqrt{n+40} < \int_n^{n+1} \sqrt{x+40}\, dx.…
  34. 34Find ∑r=0n[(n+2)r+n]\sum_{r=0}^{n} [(n+2)r + n], giving your answer in terms of nn. [You may use the result ∑r=1nr3=14n2(n+1)2\sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2…
  35. 35A data processing centre processes data (in suitable units) every day. Let unu_n, where nn is an integer, n≥1n \ge 1, re…
  36. 36A sequence u1,u2,u3,…u_1,u_2,u_3,\dots is such that u1=34u_1 = \dfrac{3}{4} and un+1=un+34(12)2nu_{n+1} = u_n + \dfrac{3}{4}\left(\dfrac{1}{2}\right)^{2n}…
  37. 37A sequence u0, u1, u2, …u_0,\ u_1,\ u_2,\ \dots is such that u0=0u_0 = 0 and un=un−1−2−n(n2−4n+2),for all n∈Z+.u_n = u_{n-1} - 2^{-n}\left(n^2 - 4n + 2\right),\quad\text{for all } n \in \mathbb{Z}^+.…
  38. 38The 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…
  39. 39An arithmetic progression has first term aa and common difference dd, where aa and dd are non-zero. The first, third…
  40. 40It is given that f(r)=1−3(r+2)!\mathrm{f}(r) = 1 - \frac{3}{(r+2)!}, where rr is a non-negative integer. Show that f(r)−f(r−1)=a(r+1)(r+2)!\mathrm{f}(r) - \mathrm{f}(r-1) = \frac{a(r+1)}{(r+2)!}…
  41. 41The nth term of a series is 2n−2+3n2^{n-2} + 3n. Find the sum of the first NN terms.
  42. 42已知Sn{S}_{n} 是公差不为00的等差数列{an}\left\{{a}_{n}\right\} 的前nn项和，S5=35{S}_{5}=35 ，a1,a4,a13{a}_{1},{a}_{4},{a}_{13} 成等比数列． (1…
  43. 43Show that 32r−1−42r+1+12r+3=8r+20(2r−1)(2r+1)(2r+3)\frac{3}{2r-1} - \frac{4}{2r+1} + \frac{1}{2r+3} = \frac{8r+20}{(2r-1)(2r+1)(2r+3)}. Hence show that ∑r=1n2r+5(2r−1)(2r+1)(2r+3)=23−n+2(2n+1)(2n+3)\sum_{r=1}^{n} \frac{2r+5}{(2r-1)(2r+1)(2r+3)} = \frac{2}{3} - \frac{n+2}{(2n+1)(2n+3)}…
  44. 44Ryan has playing cards which he stacks into pyramids. He will begin by stacking up cards to form a pyramid with 1 level,…
  45. 45Given that ∑r=1n1r(r+1)=nn+1\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}. Show that ∑r=2n1r(r−1)=1−1n\sum_{r=2}^{n} \frac{1}{r(r-1)} = 1 - \frac{1}{n}…
  46. 46A sequence of negative real numbers x1,x2,x3,…x_1, x_2, x_3, \dots satisfies the relation xn+1=−1−2xn, for n≥1. x_{n+1} = -\sqrt{1-2x_n}, \text{ for } n \ge 1.…
  47. 47A 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…
  48. 48已知等差数列{an}\{a_{n}\}的公差d>0d>0，前nn项和为SnS_{n}，且满足a2a3=45a_{2}a_{3}=45，a1+a4=14.(1)a_{1}+a_{4}=14.(1)求数列{an}\{a_{n}\}的通项公式及前nn…
  49. 49A sequence u1,u2,u3,…u_1, u_2, u_3, \dots is such that u1=13u_1 = \frac{1}{3} and ur+1=ur−1(2r−1)(2r+3),for all r≥1.u_{r+1} = u_r - \frac{1}{(2r-1)(2r+3)}, \quad \text{for all } r \ge 1.…
  50. 50It is given that ∑r=1n7r+4r(r+1)(r+2)=92−2n+1−5n+2.\sum_{r=1}^{n} \dfrac{7r+4}{r(r+1)(r+2)} = \dfrac{9}{2} - \dfrac{2}{n+1} - \dfrac{5}{n+2}. Find ∑r=72n7r−3r3−r\sum_{r=7}^{2n} \dfrac{7r-3}{r^3 - r}…
  51. 51A sequence u0,u1,u2,…u_0, u_1, u_2, \dots is such that u0=1u_0 = 1 and un=un−1+2n+1n(n+1),for all n∈N+.u_n = u_{n-1} + 2n + \frac{1}{n(n+1)}, \quad\text{for all } n \in \mathbb{N}^+.…
  52. 52It is given that f(n)=n5n−1\text{f}(n) = \frac{n}{5^{n-1}} where nn is a positive integer. By considering f(r)−f(r+1)\text{f}(r) - \text{f}(r+1)…
  53. 53It is given that ∑r=1n2r(r+1)(r+2)=12−1(n+1)(n+2).\sum_{r=1}^n \dfrac{2}{r(r+1)(r+2)} = \dfrac{1}{2} - \dfrac{1}{(n+1)(n+2)}. State the sum to infini…
  54. 54Prove by induction that, for all positive integers nn and for x∈Rx \in \mathbb{R}, x≠0,x≠1x \ne 0, x \ne 1, 1x−1−1x−1x2−⋯−1xn=1xn(x−1).\dfrac{1}{x-1}-\dfrac{1}{x}-\dfrac{1}{x^2}-\dots-\dfrac{1}{x^n}=\dfrac{1}{x^n(x-1)}.…
  55. 55Given that ∑r=1n1r(r+1)=nn+1\displaystyle\sum_{r=1}^{n} \frac{1}{r(r+1)} = \frac{n}{n+1}. Show that ∑r=2n1r(r−1)=1−1n\displaystyle\sum_{r=2}^{n} \frac{1}{r(r-1)} = 1 - \frac{1}{n}…
  56. 56A sequence of negative real numbers x1,x2,x3,…x_1,x_2,x_3,\dots satisfies the relation xn+1=−1−2xn, for n≥1.x_{n+1} = -\sqrt{1-2x_n},\text{ for } n\ge 1.…
  57. 57A sequence u1,u2,u3,…u_1, u_2, u_3, \dots is such that u1=−1u_1 = -1 and un+1=un+2(3−n)+au_{n+1} = u_n + 2(3^{-n}) + a, where n∈Z+n \in \mathbb{Z}^+…
  58. 58By writing 1(2r+1)(2r+3)\frac{1}{(2r+1)(2r+3)} in partial fractions, find an expression for ∑r=1n1(2r+1)(2r+3)\sum_{r=1}^{n} \frac{1}{(2r+1)(2r+3)}…
  59. 59A company posted a video on a social media platform to advertise a new product. The video is uploaded at the start of 1st1^{\text{st}}…
  60. 60A series is given by ∑r=1n2(4−3x)r\sum_{r=1}^{n} 2(4-3x)^{r} where xx is constant. Explain why this is a geometric series. Deter…