Prove that (a+b)/(an - bn) , if n is a positive even integer.

We prove it with the help of mathematical induction.

CASE – I

For = n=2

aⁿ – bⁿ = a² - b²

⇒ (a+b)/( a² - b²)

CASE-I is true for n=2.

CASE-II

Suppose it is true for n=k , where k is a positive even integer.

⇒(a+b)/( a^k - b^k) says (1)

We prove it for n=k+2

aⁿ – bⁿ = a^k+2 - b^k+2

= a^k . a² - b^k. b²

Adding & Subtracting a^k . b² , we have

= a^k . a² - b^k. b² + a^k . b² - a^k . b²

= a^k . a² - a^k . b² + a^k . b² - b^k. b²

= a^k (a² - b²) + b² (a^k- b^k) says (2)

We know that

⇒(a+b)/( a² - b²)

⇒(a+b)/ a^k (a² - b²) says (3)

From (1) & (3), we have

⇒(a+b)/ a^k (a² - b²) + b² (a^k- b^k)

From (2), we have

⇒(a+b)/( aⁿ- bⁿ)

CASE-II is true for n=k+2

Hence proved (a+b)/( aⁿ - bⁿ) is true for every positive even integer.

Prove that (a+b)/(an - bn) , if n is a positive even integer.

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