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Math Help - Induction question help

  1. #1
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    Induction question help

    If a1 = 2 and an = an−1 + 5, then an = 5n − 3.

    any idea how i can prove this equation? much appreciated
    Last edited by CaptainBlack; March 9th 2009 at 11:09 PM. Reason: to get rid of all the gash formating which was screwing up this post
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  2. #2
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    Sequence

    Hello jvignacio
    Quote Originally Posted by jvignacio View Post
    If a1 = 2 and an = an1 + 5, then an = 5n 3.

    any idea how i can prove this equation? much appreciated
    Although you have headed this post 'Induction' you can prove this directly, since a_1 = 2 and a_n = a_{n-1}+ 5 is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The n^{th} term is a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello jvignacioAlthough you have headed this post 'Induction' you can prove this directly, since a_1 = 2 and a_n = a_{n-1}+ 5 is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The n^{th} term is a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3

    Grandad
    ahhh yes thank you !
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Hello jvignacioAlthough you have headed this post 'Induction' you can prove this directly, since a_1 = 2 and a_n = a_{n-1}+ 5 is the definition of an arithmetic sequence, whose first term is 2, common difference 5. The n^{th} term is a_n=a + (n-1)d = 2 + 5(n-1) = 5n -3

    Grandad
    Hi Grandad, if we were to prove this by induction proof, would it be done the same way ? because i didnt see you tryed the base case first then n+1. Cheers
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  5. #5
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    Induction proof

    Hello jvignacio
    Quote Originally Posted by jvignacio View Post
    Hi Grandad, if we were to prove this by induction proof, would it be done the same way ? because i didnt see you tryed the base case first then n+1. Cheers
    No, it's completely different by induction, and it looks like this:

    Suppose that P(n) is the propositional function: a_n=5n-3

    Then, since a_{n+1}=a_n+5, P(n)\Rightarrow a_{n+1}=5n-3+5=5(n+1)-3

    So P(n) \Rightarrow P(n+1)

    Now P(1) is: a_1 = 5\times 1 -3 = 2, which is true.

    So P(n) is true for all n \ge 1.

    Do you understand what all this means?

    Grandad
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    Quote Originally Posted by Grandad View Post
    Hello jvignacioNo, it's completely different by induction, and it looks like this:

    Suppose that P(n) is the propositional function: a_n=5n-3

    Then, since a_{n+1}=a_n+5, P(n)\Rightarrow a_{n+1}=5n-3+5=5(n+1)-3

    So P(n) \Rightarrow P(n+1)

    Now P(1) is: a_1 = 5\times 1 -3 = 2, which is true.

    So P(n) is true for all n \ge 1.

    Do you understand what all this means?

    Grandad
    hrmm not really. you lost me at the beginning.. i will try study it more
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  7. #7
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    Quote Originally Posted by jvignacio View Post
    If a1 = 2 and an = an−1 + 5, then an = 5n − 3.

    any idea how i can prove this equation? much appreciated
    Proof by induction

    Lets consider this as a statement
    (this means P(h) addresses for n= h)


    Show it for n= 1


    P(1): a1= 5(1) -3 = 2

    Its given in the question that a1 =2
    Thus this statement is correct for n=1

    Assume it correct for n= k

    So we have assumed that
    P(k) is correct this means

    P(k):  a_k = 5k-3

    Using above assumption prove it for n= k+1

    So we need to prove that P(k+1) is correct using the assumption that P(k) is correct
    steps
    TO Prove
    P(k+1): a_{k+1} = 5(k+1)-3

    Given
    a_k = 5k  -3

    & a_n =a_{n-1} + 5

    Proof:

    a_{k+1} = a_{k} + 5

    <br />
a_{k+1} = 5k-3 + 5  = 5(k+1) -3

    This proves that its correct for k+1

    Hence its correct for all real numbers through the principle of mathematical induction
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  8. #8
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    Quote Originally Posted by ADARSH View Post
    Proof by induction

    Lets consider this as a statement
    (this means P(h) addresses for n= h)


    Show it for n= 1

    P(1): a1= 5(1) -3 = 2

    Its given in the question that a1 =2
    Thus this statement is correct for n=1

    Assume it correct for n= k

    So we have assumed that
    P(k) is correct this means

    P(k):  a_k = 5k-3

    Using above assumption prove it for n= k+1

    So we need to prove that P(k+1) is correct using the assumption that P(k) is correct
    steps
    TO Prove
    P(k+1): a_{k+1} = 5(k+1)-3

    Given
    a_k = 5k -3

    & a_n =a_{n-1} + 5

    Proof:

    a_{k+1} = a_{k} + 5

    <br />
a_{k+1} = 5k-3 + 5 = 5(k+1) -3

    This proves that its correct for k+1

    Hence its correct for all real numbers through the principle of mathematical induction
    thanks for that. really cleared things up well for me.

    jv
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