What is the value of \(\displaystyle\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\cdots+\frac{1}{\sqrt{80}+\sqrt{81}}\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 8
Each term is \(\frac{1}{\sqrt{k}+\sqrt{k+1}}\). Multiplying top and bottom by \(\sqrt{k+1}-\sqrt{k}\) turns the bottom into \((k+1)-k=1\), so each term equals \(\sqrt{k+1}-\sqrt{k}\) and in the sum almost every root cancels.
\(\displaystyle\frac{1}{\sqrt{k}+\sqrt{k+1}}=\frac{\sqrt{k+1}-\sqrt{k}}{(k+1)-k}=\sqrt{k+1}-\sqrt{k}\)
Sum \(=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+(\sqrt{4}-\sqrt{3})+\cdots+(\sqrt{81}-\sqrt{80})\)
Only the last and the first parts remain: \(\sqrt{81}-1\)
\(\sqrt{81}-1=9-1=8\)