What is the value of \(\frac{1}{1+\sqrt{2}}+ \frac{1}{\sqrt{2} + \sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}?\)
9
Solution :
S = Σ 1/(√n + √(n+1)), where n = 1 to 99
1/(√n + √(n+1))
= [1/(√n + √(n+1))] × [(√(n+1) − √n)/(√(n+1) − √n)]
= (√(n+1) − √n) / [(n+1) − n]
= √(n+1) − √n
S = (√2 − 1) + (√3 − √2) + (√4 − √3) + ... + (√100 − √99)
All middle terms cancel each other.
S = √100 − 1
= 10 − 1
= 9
9
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