Huge Summation!

Algebra Level 4

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


The answer is 9.

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1 solution

Aareyan Manzoor
Dec 4, 2015

write as 1 ( n 4 + n + 1 4 ) ( n + n + 1 ) = ( n 4 − n + 1 4 ) ( n 4 − n + 1 4 ) ( n 4 + n + 1 4 ) ( n + n + 1 ) \dfrac{1}{(\sqrt[4]{n}+\sqrt[4]{n+1})(\sqrt{n}+\sqrt{n+1})}=\dfrac{(\sqrt[4]{n}-\sqrt[4]{n+1})}{(\sqrt[4]{n}-\sqrt[4]{n+1})(\sqrt[4]{n}+\sqrt[4]{n+1})(\sqrt{n}+\sqrt{n+1})} = ( n 4 − n + 1 4 ) ( n − n + 1 ) ( n + n + 1 ) = − ( n 4 − n + 1 4 ) = n + 1 4 − n 4 =\dfrac{(\sqrt[4]{n}-\sqrt[4]{n+1})}{(\sqrt{n}-\sqrt{n+1})(\sqrt{n}+\sqrt{n+1})}=-(\sqrt[4]{n}-\sqrt[4]{n+1})=\sqrt[4]{n+1}-\sqrt[4]{n} so ∑ n = 1 9999 ( n + 1 4 − n 4 ) = 9999 + 1 4 − 1 4 = 9 \sum_{n=1}^{9999} (\sqrt[4]{n+1}-\sqrt[4]{n})=\sqrt[4]{9999+1}-\sqrt[4]{1}=\boxed{9} if you are not familiar with telescoping series, google it.

8 Helpful 0 Interesting 0 Brilliant 0 Confused

Same way. I kept forgetting the negative sign but got it right in my last try

Shreyash Rai - 5 years, 6 months ago

I did the same

Aditya Kumar - 5 years, 1 month ago

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