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Mathematics

Find the value of :

11+2+12+3+...+12025+2026\dfrac{1}{1 + \sqrt{2}} + \dfrac{1}{\sqrt{2} + \sqrt{3}} + …+ \dfrac{1}{\sqrt{2025} + \sqrt{2026}}

Rational Irrational Nos

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Answer

Rationalizing a general term,

1n+n+1\frac{1}{\sqrt{n} + \sqrt{n+1}}

To rationalize, we multiply the numerator and denominator by the conjugate, (n+1−n)(\sqrt{n+1} - \sqrt{n}):

1n+1+n×n+1−nn+1−n=n+1−n(n+1)2−(n)2=n+1−n(n+1)−n=n+1−n\dfrac{1}{\sqrt{n+1} + \sqrt{n}} \times \dfrac{\sqrt{n+1} - \sqrt{n}}{\sqrt{n+1} - \sqrt{n}} \\[1em] = \dfrac{\sqrt{n+1} - \sqrt{n}}{(\sqrt{n+1})^2 - (\sqrt{n})^2} \\[1em] = \dfrac{\sqrt{n+1} - \sqrt{n}}{(n+1) - n} \\[1em] = \sqrt{n+1} - \sqrt{n}

Expanding the series,

First term: 11+2=2−1\dfrac{1}{1 + \sqrt{2}} = \sqrt{2} - 1

Second term: 12+3=3−2\dfrac{1}{\sqrt{2} + \sqrt{3}} = \sqrt{3} - \sqrt{2}

Third term: 13+4=4−3\dfrac{1}{\sqrt{3} + \sqrt{4}} = \sqrt{4} - \sqrt{3}

…

Last term: 12025+2026=2026−2025\dfrac{1}{\sqrt{2025} + \sqrt{2026}} = \sqrt{2026} - \sqrt{2025}

Adding all the series,

(2−1)+(3−2)+(4−3)+⋯+(2026−2025)(\sqrt{2} - 1) + (\sqrt{3} - \sqrt{2}) + (\sqrt{4} - \sqrt{3}) + \dots + (\sqrt{2026} - \sqrt{2025})

2−1+3−2+4−3+⋯+2026−2025\sqrt{2} - 1 + \sqrt{3} - \sqrt{2} + \sqrt{4} - \sqrt{3} + \dots + \sqrt{2026} - \sqrt{2025}

Everything cancels except the first and last term.

-1 + 2026\sqrt{2026}.

Hence, solution = 2026\sqrt{2026} - 1.

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