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secondary 3 | A Maths
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How to do part (ii)
Date Posted: 4 years ago
Views: 225
Hence means that you are expected to use part i)'s result to do part ii)
You have already shown that :
1/(√x+1 + √x) = √x+1 - √x
So,
1/(√100 + √99) + 1/(√99 + 98) + ... + 1/(√98 + 97) + ... 1/(√3 + √2) + 1/(√2 + √1)
= √100 - √99 + √99 - √98 + ... + √3 - √2 + √2 - √1
= √100 - √1
= 10 - 1
= 9
(Realise that most of the terms cancel out,
eg. -√99 + √99 gives you 0, -√98 + √98, -√3 + √3, -√2 + √2 all gives you 0.
You're only left with √100 and √1 not cancelled out. From there it is simple to solve)
1/(√x+1 + √x) = √x+1 - √x
So,
1/(√100 + √99) + 1/(√99 + 98) + ... + 1/(√98 + 97) + ... 1/(√3 + √2) + 1/(√2 + √1)
= √100 - √99 + √99 - √98 + ... + √3 - √2 + √2 - √1
= √100 - √1
= 10 - 1
= 9
(Realise that most of the terms cancel out,
eg. -√99 + √99 gives you 0, -√98 + √98, -√3 + √3, -√2 + √2 all gives you 0.
You're only left with √100 and √1 not cancelled out. From there it is simple to solve)
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This is a useful technique to know, called the method of differences.
Date Posted:
4 years ago
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