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解答 - 几何数列

公比是：

r = 0.12452830188679245

r=0.12452830188679245

该系列的和是：

s = 596

s=596

此系列的通用形式是：

a_{n} = 530 \cdot {0.12452830188679245}^{n - 1}

a_n=530*0.12452830188679245^(n-1)

这个序列的第n项是：

530, 66, 8.218867924528302, 1.0234816660733357, 0.12745243388837765, 0.015871435163458347, 0.0019764428694117944, 0.00024612307430411023, 3.0649288498247695 E - 05, 3.816703850725184 E - 06

530,66,8.218867924528302,1.0234816660733357,0.12745243388837765,0.015871435163458347,0.0019764428694117944,0.00024612307430411023,3.0649288498247695E-05,3.816703850725184E-06

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：

$a_{2} a_{1} = \frac{66}{530} = 0.12452830188679245$

该序列的公比（ $r$ ）保持不变，并且等于两个连续项的商。
$r = 0.12452830188679245$

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： $a = 530$ 、公比： $r = 0.12452830188679245$ 和元素数目 $n = 2$ 插入几何级数求和公式：

$s_{2} = 530 * ((1 - {0.12452830188679245}^{2}) / (1 - 0.12452830188679245))$

$s_{2} = 530 * ((1 - 0.015507297970808116) / (1 - 0.12452830188679245))$

$s_{2} = 530 * (0.9844927020291919 / (1 - 0.12452830188679245))$

$s_{2} = 530 * (0.9844927020291919 / 0.8754716981132076)$

$s_{2} = 530 \cdot 1.1245283018867924$

$s_{2} = 596$

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： $a = 530$ 和公比： $r = 0.12452830188679245$ 插入几何级数的公式：

$a_{n} = 530 \cdot {0.12452830188679245}^{n - 1}$

4. 找到第n项

使用通用公式找到第n项

$a_{1} = 530$

$a_{2} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{2 - 1} = 530 \cdot {0.12452830188679245}^{1} = 530 \cdot 0.12452830188679245 = 66$

$a_{3} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{3 - 1} = 530 \cdot {0.12452830188679245}^{2} = 530 \cdot 0.015507297970808116 = 8.218867924528302$

$a_{4} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{4 - 1} = 530 \cdot {0.12452830188679245}^{3} = 530 \cdot 0.0019310974831572372 = 1.0234816660733357$

$a_{5} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{5 - 1} = 530 \cdot {0.12452830188679245}^{4} = 530 \cdot 0.00024047629035542953 = 0.12745243388837765$

$a_{6} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{6 - 1} = 530 \cdot {0.12452830188679245}^{5} = 530 \cdot 2.9946104081996884 E - 05 = 0.015871435163458347$

$a_{7} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{7 - 1} = 530 \cdot {0.12452830188679245}^{6} = 530 \cdot 3.729137489456216 E - 06 = 0.0019764428694117944$

$a_{8} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{8 - 1} = 530 \cdot {0.12452830188679245}^{7} = 530 \cdot 4.6438315906435894 E - 07 = 0.00024612307430411023$

$a_{9} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{9 - 1} = 530 \cdot {0.12452830188679245}^{8} = 530 \cdot 5.7828846223108854 E - 08 = 3.0649288498247695 E - 05$

$a_{10} = a_{1} \cdot r^{n - 1} = 530 \cdot {0.12452830188679245}^{10 - 1} = 530 \cdot {0.12452830188679245}^{9} = 530 \cdot 7.201328020236197 E - 09 = 3.816703850725184 E - 06$

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