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Let `A_n+1 = (1-alpha) (1-A_n)+A_n` for `n=1, 2, 3, …………………` and `A_1 =...

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Let `A_n+1 = (1-alpha) (1-A_n)+A_n` for `n=1, 2, 3, …………………` and `A_1 = beta` , Where `alpha` and `beta` are real numbers. Prove, by the Principle of Mathematical Induction, that for every positive integer n,

`A_n = 1-(1-beta)alpha^(n-1)`

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Posted (Answer #1)

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When n=1,

`A_n = 1-(1-beta)alpha^(n-1)`

`A_1 = 1-(1-beta)alpha^(1-1) = 1-(1-beta)=beta`

According to the given data this is true.

Assume that the result is true when n=p,


When n = p+1

`A_(p+1) `

`= (1-alpha)(1-beta)alpha^(p-1)+A_p`

`= (1-alpha)(1-beta)alpha^(p-1)+1-(1-beta)alpha^(p-1)`

`= (1-beta)alpha^(p-1)[(1-alpha)-1]+1`

`= (1-beta)alpha^(p-1)-alpha+1`

`= 1-(1-beta)alpha^p`

`= 1-(1-beta)alpha^(p+1)-1`

So for n = p+1 the result is true.

So using mathematical induction for every integer n>0 the result is true.


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Posted (Answer #2)

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Given a recurrence relation


We wish to prove  P(n) statement ,by Mathematical induction.

`P(n):A_n=1-(1-beta)alpha^(n-1), AAninZ^+`

Let n=1,2,3  from recurrence relation

`A_1=beta`    (given)



`=1-alpha-beta+alpha beta+beta`







Let P(k) is true and k>3.


We wish to prove P(k+1) is true when P(k) is true.








Thus P(k+1) is true when P(k) is true.

By principle of mathematical induction  P(n) is true for all n ,n is positive integers.

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