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Question

What is the value of \(\frac{1}{1+\sqrt{2}}+ \frac{1}{\sqrt{2} + \sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}?\)

This question was previously asked in
CDS I 2020 Elementary Mathematics Previous Year Paper (02-Feb-2020)
The correct answer is

9

Solution :

Step 1: Write the general term

S = Σ 1/(√n + √(n+1)),   where n = 1 to 99

Step 2: Rationalize the denominator

1/(√n + √(n+1))

= [1/(√n + √(n+1))] × [(√(n+1) − √n)/(√(n+1) − √n)]

= (√(n+1) − √n) / [(n+1) − n]

= √(n+1) − √n

Step 3: Substitute in the series

S = (√2 − 1) + (√3 − √2) + (√4 − √3) + ... + (√100 − √99)

All middle terms cancel each other.

S = √100 − 1

= 10 − 1

= 9

Final Answer:

9

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