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Math Help - [SOLVED] recursive def. Check my answer please.

  1. #1
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    [SOLVED] recursive def. Check my answer please.

    Find f(2), f(3), f(4), f(5) if f(n) is defined recursively by f(0)= 0, f(1) = 1, f(n+1) = f(n)+2f(n-1)+1 for n = 1, 2, ....

    f(2) = f(1) +2(1-1)+1
    f(2) = 2

    f(3) = f(2) + 2(2-1)+1
    f(3) = 5

    f(4) = f(3) + 2(5-1)+1
    f(4) = 14

    f(5) = f(4) + 2(14-1)+1
    f(5) = 41

    correct?
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  2. #2
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    Quote Originally Posted by ninobrn99 View Post
    Find f(2), f(3), f(4), f(5) if f(n) is defined recursively by f(0)= 0, f(1) = 1, f(n+1) = f(n)+2f(n-1)+1 for n = 1, 2, ....

    f(2) = f(1) +2(1-1)+1
    f(2) = 2

    f(3) = f(2) + 2(2-1)+1
    f(3) = 5

    f(4) = f(3) + 2(5-1)+1
    f(4) = 14

    f(5) = f(4) + 2(14-1)+1
    f(5) = 41

    correct?
    the formula is f( n + 1 ) = f( n ) + 2*f( n - 1 ) + 1
    and you are making mistake because you did:
    f( n + 1 ) = f( n ) +2*( n - 1 ) + 1

    answers:

    f( 2 ) = f( 1 ) + 2*f( 1 - 1 ) + 1 = 1 + 2*0 + 1 = 2

    f( 3 ) = f( 2 ) + 2*f( 1 ) + 1 = 2 + 2*1 + 1 = 5

    f( 4 ) = f( 3 ) + 2*f( 2 ) + 1 = 5 + 2*2 + 1 = 10

    f( 5 ) = f( 4 ) + 2*f( 3 ) + 1 = 10 + 2*5 + 1 = 21
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  3. #3
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    Quote Originally Posted by josipive View Post
    the formula is f( n + 1 ) = f( n ) + 2*f( n - 1 ) + 1
    and you are making mistake because you did:
    f( n + 1 ) = f( n ) +2*( n - 1 ) + 1

    answers:

    f( 2 ) = f( 1 ) + 2*f( 1 - 1 ) + 1 = 1 + 2*0 + 1 = 2

    f( 3 ) = f( 2 ) + 2*f( 1 ) + 1 = 2 + 2*1 + 1 = 5

    f( 4 ) = f( 3 ) + 2*f( 2 ) + 1 = 5 + 2*2 + 1 = 10

    f( 5 ) = f( 4 ) + 2*f( 3 ) + 1 = 10 + 2*5 + 1 = 21
    Thanks for showing me what I was doing wrong! It helped tremendously!
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