Find the sum of ‘n’ terms of the series : -Maths 9th

Find the sum of ‘n’ terms of the series : -Maths 9th
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(b) \(n\, ext{log}_2\big(rac{x}{y}\big)\)Given series=  \( ext{log}_2\big(rac{x}{y}\big)\) + \( ext{log}_{2^2}\big(rac{x}{y}\big)^2\) + \( ext{log}_{2^3}\big(rac{x}{y}\big)^3\) + \( ext{log}_{2^4}\big(rac{x}{y}\big)^4\) + .............= \( ext{log}_2\big(rac{x}{y}\big)\) + \( ext{log}_2\big(rac{x}{y}\big)\) + \( ext{log}_2\big(rac{x}{y}\big)\) + \( ext{log}_2\big(rac{x}{y}\big)\) + .......n terms= \( ext{log}_2\big(rac{x}{y}.rac{x}{y}.rac{x}{y}.rac{x}{y}.\,.......n \, ext{terms}\big)\)= \( ext{log}_2\big(rac{x}{y}\big)^n\) = \(n\, ext{log}_2\big(rac{x}{y}\big)\).
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Last Answer : 5√2 + 20√2 = (5+20)√2 = 25√2

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Last Answer : Reducing into simplest form √12 = √22 × 3 = 2√3 √75 = √3 × 52 = 5√3 Therefore 4√3 - 3√12 + 2√75 = 4√3 - 3 × 2√3 + 2 × 5√3 = 4√3 - 6√3 + 10√3 = (4 - 6 + 10) √3 = 8√3

Simplify by combining similar terms : -Maths 9th

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Last Answer : Reducing into simplest form √8 = √(22× 2) = 2 × √2 √32 = √(22 × 22 × 2) = 2 × 2√2 = 4√2 ∴ √8 + √32 - √2 = 2√2 + 4√2 - √2 = (2 + 4 -1) √2 = 5√2 .

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Last Answer : Reducing into simplest form : √45 = √32 × 5 = 3√5 √20 = √22 × 5 = 2√5 ∴ √45 -3√20 + 4√5 = 3√5 - 3 × 2√5 + 4√5 = 3√5 - 6√5 + 4√5 = √5

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Last Answer : √12 = √(22 × 3) = 2√3 √50 = √(2 × 52) = (5 √2) √48 = √(22 × 22 × 3) = (2 × 2√3)= 4√3 ∴ 4√12 - √50 - 7 √48 = 4 × 2√3 - 5√2 - 7 × 4√3 = 8√3 - 5√2 - 28√3 = (8 -28)√3 - 5√2 = -20√3 - 5√2

Simplifying by combining similar terms: -Maths 9th

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Last Answer : Reducing into simplest form

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