series log a+loga^2/b+log a^3/b^2+-…ke n pdo ka yog gyat kijiye ​

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series log a+loga^2/b+log a^3/b^2+…….ke n pdo ka yog gyat kijiye

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Clara 3 weeks 2021-10-01T14:59:47+00:00 1 Answer 0 views 0

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    2021-10-01T15:01:31+00:00

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    Cᴏʀʀᴇᴄᴛ Qᴜᴇꜱᴛɪᴏɴ :

    Find the sum of the series

    log(a) + log[a²/b] + log[a³/b²] + ……… + n

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    Sᴏʟᴜᴛɪᴏɴ :

    Series :

    ➨ log(a) + log[a²/b] + log[a³/b²] …… + n

    It can be written as

    ➨ log[a/b⁰] + log[a²/b] + log[a³/b²] …… + n

    Simplifying by using the formula

    log(a) + log(b) = log(ab)

    Similarly

    ➨ log[ a .a².a³.…..aⁿ/b⁰.b.b².b³….bⁿ⁻¹ ]

    Usi ng

    aⁿ × a = a

    ➨ log[ a¹⁺²⁺³⁺…⁺ⁿ/b⁰⁺¹⁺²⁺³⁺…⁺ⁿ⁻¹ ]

    As we know that

    Sum of n natural numbers = n(n + 1)/2

    And for n 1 , substituting n = n 1

    Sum of n-1 natural numbers = (n 1)(n 1 + 1)/2 = n(n 1)/2

    ➨ log{ a^[ n(n + 1)/2 ]/{b^[ n(n – 1)/2 ] }

    Taking common in the powers

    ➨ log{ [ aⁿ⁺¹/aⁿ⁻¹ ]^(n/2) }

    Now , using the formula

    log aⁿ = n log(a)

    n/2 log[ aⁿ⁺¹/aⁿ⁻¹ ]

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    Hence , log(a) + log[a²/b] + log[a³/b²] + ……… + n = n/2 log[ aⁿ⁺¹/aⁿ⁻¹ ]

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