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# MATHS PROBLEM

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

=? 5該我要答案&詳細解釋

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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

To solve:

Method:

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

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

S(r) = {&amp;radic;r - &amp;radic;(r+1) } &amp;divide; { r - (r+1) }

therefore:

or

Similarly,

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

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

So

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