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解答 - 几何数列

公比是：

r = 0.006535947712418301

r=0.006535947712418301

该系列的和是：

s = 2926

s=2926

此系列的通用形式是：

a_{n} = 2907 \cdot {0.006535947712418301}^{n - 1}

a_n=2907*0.006535947712418301^(n-1)

这个序列的第n项是：

2907, 19, 0.12418300653594773, 0.0008116536374898545, 5.304925735227807 E - 06, 3.467271722371116 E - 08, 2.2661906682164153 E - 10, 1.481170371383278 E - 12, 9.680852100544302 E - 15, 6.327354314081243 E - 17

2907,19,0.12418300653594773,0.0008116536374898545,5.304925735227807E-06,3.467271722371116E-08,2.2661906682164153E-10,1.481170371383278E-12,9.680852100544302E-15,6.327354314081243E-17

查看步骤

逐步解答

1. 找到公比

通过将序列中的任何项除以前一项来找到公比：

$a_{2} a_{1} = \frac{19}{2907} = 0.006535947712418301$

该序列的公比（ $r$ ）保持不变，并且等于两个连续项的商。
$r = 0.006535947712418301$

2. 求和

5 个额外步骤

$s_{n} = a * ((1 - r^{n}) / (1 - r))$

要找到系列的和，将第一项： $a = 2, 907$ 、公比： $r = 0.006535947712418301$ 和元素数目 $n = 2$ 插入几何级数求和公式：

$s_{2} = 2907 * ((1 - {0.006535947712418301}^{2}) / (1 - 0.006535947712418301))$

$s_{2} = 2907 * ((1 - 4.271861249946602 E - 05) / (1 - 0.006535947712418301))$

$s_{2} = 2907 * (0.9999572813875005 / (1 - 0.006535947712418301))$

$s_{2} = 2907 * (0.9999572813875005 / 0.9934640522875817)$

$s_{2} = 2907 \cdot 1.0065359477124183$

$s_{2} = 2926$

3. 找到通用形式

$a_{n} = a \cdot r^{n - 1}$

要找到系列的通用形式，将第一项： $a = 2, 907$ 和公比： $r = 0.006535947712418301$ 插入几何级数的公式：

$a_{n} = 2907 \cdot {0.006535947712418301}^{n - 1}$

4. 找到第n项

使用通用公式找到第n项

$a_{1} = 2907$

$a_{2} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{2 - 1} = 2907 \cdot {0.006535947712418301}^{1} = 2907 \cdot 0.006535947712418301 = 19$

$a_{3} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{3 - 1} = 2907 \cdot {0.006535947712418301}^{2} = 2907 \cdot 4.271861249946602 E - 05 = 0.12418300653594773$

$a_{4} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{4 - 1} = 2907 \cdot {0.006535947712418301}^{3} = 2907 \cdot 2.792066176435688 E - 07 = 0.0008116536374898545$

$a_{5} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{5 - 1} = 2907 \cdot {0.006535947712418301}^{4} = 2907 \cdot 1.8248798538795345 E - 09 = 5.304925735227807 E - 06$

$a_{6} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{6 - 1} = 2907 \cdot {0.006535947712418301}^{5} = 2907 \cdot 1.1927319306402187 E - 11 = 3.467271722371116 E - 08$

$a_{7} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{7 - 1} = 2907 \cdot {0.006535947712418301}^{6} = 2907 \cdot 7.7956335335962 E - 14 = 2.2661906682164153 E - 10$

$a_{8} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{8 - 1} = 2907 \cdot {0.006535947712418301}^{7} = 2907 \cdot 5.095185316075948 E - 16 = 1.481170371383278 E - 12$

$a_{9} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{9 - 1} = 2907 \cdot {0.006535947712418301}^{8} = 2907 \cdot 3.3301864810953912 E - 18 = 9.680852100544302 E - 15$

$a_{10} = a_{1} \cdot r^{n - 1} = 2907 \cdot {0.006535947712418301}^{10 - 1} = 2907 \cdot {0.006535947712418301}^{9} = 2907 \cdot 2.176592471304177 E - 20 = 6.327354314081243 E - 17$

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几何数列