The great sum

Algebra Level 3

n = 1 9999 1 ( n + n + 1 ) ( n 4 + n + 1 4 ) = ? \large \displaystyle\sum_{n=1}^{9999}\frac{1}{(\sqrt{n}+\sqrt{n+1})(\sqrt[4]{n}+\sqrt[4]{n+1})}= \ ?


The answer is 9.

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1 solution

Prakhar Gupta
Jun 25, 2015

This is a classic example from Telescoping Series problem.

It can be easily seen that the given expression can be quite simplified if we multiply numerator and denominator by n + 1 4 n 4 \sqrt[4]{n+1} - \sqrt[4]{n} .

Doing this we get:- n = 1 9999 n + 1 4 n 4 ( n + 1 4 n 4 ) ( n + 1 4 + n 4 ) ( n + 1 + n ) \sum_{n=1}^{9999} \dfrac{ \sqrt[4]{n+1} - \sqrt[4]{n}}{(\sqrt[4]{n+1} - \sqrt[4]{n})(\sqrt[4]{n+1}+\sqrt[4]{n})(\sqrt{n+1}+\sqrt{n})} = n = 1 9999 n + 1 4 n 4 ( n + 1 n ) ( n + 1 + n ) =\sum_{n=1}^{9999} \dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})} = n = 1 9999 n + 1 4 n 4 n + 1 n =\sum_{n=1}^{9999} \dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{n+1-n} = n = 1 9999 n + 1 4 n 4 = \sum_{n=1}^{9999} \sqrt[4]{n+1} -\sqrt[4]{n} = 2 4 1 + 3 4 2 4 + 10000 4 9999 4 = \sqrt[4]{2}-1+\sqrt[4]{3} - \sqrt[4]{2}+ \ldots \sqrt[4]{10000} - \sqrt[4]{9999} = 1 + 10 = -1+10 = 9 \boxed{=9}

good solution :)

Hamza Omar - 5 years, 11 months ago

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