Anonymous
Anonymous asked in 科學及數學數學 · 1 decade ago

MATHS PROBLEM

1÷(√1+√2)+1÷(√2+√3)+1÷(√3+√4)+......+1÷(√2008+√2009)

=? 5該我要答案&詳細解釋

2 Answers

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  • 1 decade ago
    Favorite Answer

    1÷(√1+√2)+1÷(√2+√3)+1÷(√3+√4)+......+1÷(√2008+√2009)

    Consider

    1÷(√1+√2)

    =[1÷(√1+√2)] [ (√1-√2) ÷ (√1-√2) ]

    = (√1-√2) ÷ [(√1)^2 - (√2)^2 ]

    = (√1-√2) ÷ (1-2)

    = - (√1-√2)

    Similarly,

    1÷(√2+√3)

    = -(√2-√3)

    and

    1÷(√3+√4)

    = - (√3-√4)

    ...

    1÷(√1+√2)+1÷(√2+√3)+1÷(√3+√4)+......+1÷(√2008+√2009)

    = [- (√1-√2)] + [-(√2-√3)] + [-(√2-√3)] + [- (√3-√4)]+...+ [ -(√2008-√2009)]

    = - [√1-√2+√2-√3+√2-√3+√3-√4+...+√2008-√2009]

    = -[√1-√2009]

    = -1+√2009

  • 1 decade ago

    To solve:

    1÷(√1+√2)+1÷(√2+√3)+1÷(√3+√4)+......+1÷(√2008+√2009)=?

    Method:

    Let S = [ 1÷(√1+√2) + 1÷(√2+√3) + 1÷(√3+√4) + ... + 1÷(√2008+√2009) ]

    or S = [ S(1) + S(2) + S(3) + ... + S(2008) ]

    where S(r) = 1÷[√r+√(r+1)]

    Consider the (r)th term of S, or S(r),

    S(r) = 1÷[√r+√(r+1)]

    S(r) = 1{ √r - √(r+1) } ÷ [√r+√(r+1)]{ √r - √(r+1) }

    S(r) = {√r - √(r+1) } ÷ { r - (r+1) }

    S(r) = {√r - √(r+1) } ÷ {-1}

    S(r) = - [ √r - √(r+1) ]

    S(r) = √(r+1) - √r

    therefore:

    S(1) = √(1+1) - √1 , and S(2008) = √(2008+1) - √2008

    or

    S(2008) = √2009 - √2008 , and S(1) = √2 - √1

    Similarly,

    S(2) = √3 - √2

    S(3) = √4 - √3

    S(2006) = √2007 - √2006

    S(2007) = √2008 - √2007

    Since S = S(1) + S(2) + S(3) + ... + S(2008)

    S = S(2008) + S(2007)+ S(2006) + ... + S(3) + S(2) + S(1)

    S = ( √2009 - √2008 ) + ( √2008 - √2007 ) + ( √2007 - √2006) + ... + ( √4 - √3 ) + ( √3 - √2 ) + ( √2 - √1)

    So

    S = √2009 - 1

    2009-03-03 00:21:45 補充:

    Sorry, I don't know why the font is messed up.

    2009-03-03 00:41:02 補充:

    Let S = [ S(1) + S(2) + S(3) + ... + S(2008) ]

    where S(r) = sqrt(r+1) - sqrt(r) and r belongs to 1, 2, through 2007 & 2008

    or

    S = S(2008) + S(2007) + S(2006) + ... + S(3) + S(2) + S(1)

    S = [ sqrt(2009) - sqrt(2008) ] + [ sqrt(2008) - sqrt(2007) ] + ... + [ sqrt(2) - sqrt(1) ]

    Anyway,

    S = sqrt(2009) - 1

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