Question

the sum 1/(√2+1) + 1/(√3+√2) + 1/(√4+√3) + ... 1/(√100+√99)

Answer 1

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║║║║  ANSWER║║║

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You should rationalise each term by multiplying them with their conjugated.

For example, (1/(√1+√2))*((√2–√1)/(√2–√1))

Which gives out √2 - √1

Second term -> √3 - √2

Third term -> √4 - √3

………goes on………

Last term -> √100 - √99

Series:√2 - √1 + √3 - √2 + √4 - √3 .…..+ √99 - √98 + √100 -√99

Observe that except (-√1) and (√100) , all the terms cancel out.

The answer is -√1 + √100 = -1 + 10 = 9