Question

series log a+loga^2/b+log a^3/b^2+.......ke n pdo ka yog gyat kijiye
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Answer 1

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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 × aᵐ = aⁿ⁺ᵐ

➨ 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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