高级数学 | 高一下册 | 第十六章 | 16.5 简易特殊数列的和 The Sum of Some Simple Progression - 独中课程 | 线上补习

高中一 | 高级数学

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16.5 The Sum of Some Simple Progression 简易特殊数列的和

Calculate the sum of the following series

计算下列各级数的和

2 x (-2) + 5 x 2 + 10 x 6 + 17 x 10 + . . . + (n $^2$ + 1)(4n – 6)

Answer:

n $^4$ – 5n

1 x 1 + (-1) x 5 + (-3) x 9 + (-5) x 13 + …. + (-37) x 77

Answer:

-19360

2 $^2$ x 2 + 3 $^2$ x 5 + 4 $^2$ x 8 + 5 $^2$ x 11 + … + 11 $^2$ x 29

Answer:

11045

1 x 4 x 7 + 2 x 5 x 8 + 3 x 6 x 9 + … + 10 x 13 x 16

Answer:

7480

18 $^2$ – 17 $^2$ + 16 $^2$ – 15 $^2$ + 14 $^2$ – 13 $^2$ + … + 2 $^2$ – 1 $^2$

Answer:

171

1 + $\frac{4}{5}$ + $\frac{7}{25}$ + $\frac{10}{125}$ + …

Answer:

$\frac{35}{16}$

1 + $\frac{3}{3}$ + $\frac{7}{9}$ + $\frac{15}{27}$ + $\cdots$

Answer:

$4\frac{1}{2}$

1 + $\frac{5}{2}$ + $\frac{9}{4}$ + $\frac{13}{8}$ + $\cdots$ to n term（至第n项）

Answer:

10 – $\frac{1}{2^{n-4}}$ – $\frac{4n-3}{2^{n-1}}$

1 + (1 + a)r + (1 + a + a $^2$ )r $^2$ + (1 + a + a $^2$ + a $^3$ )r $^3$ + … where（其中） |ar| < 1.

Answer:

$\frac{1}{(1-ar)(1-r)}$

10.

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Answer:

$\frac{55}{16}$

11.

$\frac{1}{2\times3}$ + $\frac{1}{3\times4}$ + $\frac{1}{4\times5}$ + $\frac{1}{5\times6}$ + $\frac{1}{6\times7}$ + $\cdot\cdot\cdot$ + $\frac{1}{51\times52}$

Answer:

$\frac{25}{52}$

12.

$\frac{1}{2\cdot4}$ + $\frac{1}{4\cdot6}$ + $\frac{1}{6\cdot8}$ + $\cdot\cdot\cdot$

Answer:

$\frac{1}{4}$

13.

$\sum \limits_{i=1}^{n}\frac{1}{(2i-1)(2i+1)}$

Answer:

$\frac{n}{2n+1}$

14.

$\frac{1}{1\times4}$ + $\frac{1}{2\times5}$ + $\frac{1}{3\times6}$ + $\cdot\cdot\cdot$ + $\frac{1}{20\times23}$

Answer:

$\frac{119}{253}$

15.

Simplify $\sum$ (3n – 2) $^2$ . Hence, find the value of 732 + 762 + 792 + … + 2082 .
化简 $\sum$ (3n – 2) $^2$ . 据此, 求 732 + 762 + 792 + … + 2082 的值。

Answer:

$3n^3$ – $\frac{3}{2}n^2$ – $\frac{1}{2}n$ , 981019

16.

Define a series such that, the first term is 3, the second term is 3 + 8, the third term is 3 + 8 + 13, the fourth term is 3 + 8 + 13 + 18 and so on. Find the sum of the first n terms of this series.

一级数其首项是3, 第二项是 3 + 8, 第三项是 3 + 8 + 13, 第四项是 3 + 8 + 13 + 18 以此类推。求这个级数首n 项的和。

Answer:

$\frac{n(n+1)(10n+8)}{12}$