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等比数列の和の公式の導出です。
LibreOffice 数式(Math) のソース:
sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = a + a cdot r + a cdot r ^2 + a cdot r ^3 + dotsaxis + a cdot r ^ { n - 1 }
r cdot sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = a cdot r + a cdot r ^2 + a cdot r ^3 + dotsaxis + a cdot r ^ { n - 1 } + a cdot r ^ { n }
alignl sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } - r cdot sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = left ( a + a cdot r + a cdot r ^2 + a cdot r ^3 + dotsaxis + a cdot r ^ { n - 1 } right ) -
left ( a cdot r + a cdot r ^2 + a cdot r ^3 + dotsaxis + a cdot r ^ { n - 1 } + a cdot r ^ { n } right )
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alignl phantom { y } = a - a cdot r ^ { n }
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alignl dlrarrow ( 1 - r ) cdot sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = a left ( 1 - r ^ { n } right )
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alignl dlrarrow sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = { alignc { a left ( 1 - r ^ n right ) } over { 1 - r } }
left ( a cdot r + a cdot r ^2 + a cdot r ^3 + dotsaxis + a cdot r ^ { n - 1 } + a cdot r ^ { n } right )
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alignl phantom { y } = a - a cdot r ^ { n }
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alignl dlrarrow ( 1 - r ) cdot sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = a left ( 1 - r ^ { n } right )
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alignl dlrarrow sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = { alignc { a left ( 1 - r ^ n right ) } over { 1 - r } }
alignl sum from { k = 1 } to { n } { a cdot r ^ { k - 1 } } = a + a cdot r + a cdot r ^2 + a cdot r ^3 + dotsaxis + a cdot r ^ { n - 1 } = { alignc { a left ( 1 - r ^ n right ) } over { 1 - r } }