MATHS PROBLEM

Question

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

=? 5該我要答案&詳細解釋

Answer 1

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

Answer 2

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