What is {1}{1+{2}}+{1}{{2}+{3}}+{1}{{3}+{4}}++{1}{{2020}+{2021}} equal to?

Correct Answer is : 2021‾−1{{2021}-1}2021−1

Correct Answer is : 2021‾−1{{2021}-1}2021−1

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UPSC CDS 1 2022 Math Paper

Question : 45 of 100

Marks: +1, -0

What is

\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots+\frac{1}{\sqrt{2020}+\sqrt{2021}}

equal to?

Solution: 👈: Video Solution

Given:

\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+\dots+\frac{1}{\sqrt{2020}+\sqrt{2021}}

Concept used:

(a+b)(a-b)=a^{2}-b^{2}

Calculation:

Consider the first term only, i.e.

\frac{1}{1+\sqrt{2}}

Rationalize the denominator

\Rightarrow \frac{1}{1+\sqrt{2}} \times \frac{\sqrt{2}-1}{\sqrt{2}-1}

\Rightarrow \frac{\sqrt{2}-1}{(\sqrt{2})^{2}-(1)^{2}}

\Rightarrow \sqrt{} \overline{2}-1

After doing a similar operation in other terms, we will get

\sqrt{2}-1+\sqrt{\overline{3}}-\sqrt{\overline{2}}+\sqrt{\overline{4}}-\sqrt{\overline{3}} \dots +\sqrt{2021}-\sqrt{2020}

\Rightarrow \sqrt{2021}-1

\therefore

Required value of the expression is

\sqrt{2021}-1

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