Find sum of the series - 2

Algebra Level 3

s = 1 + 1 2 3 + 1 3 3 + 1 4 3 + 1 5 3 + + 1 100 3 s=1+\frac 1{\sqrt[3]{2}}+\frac 1{\sqrt[3]{3}}+\frac 1{\sqrt[3]{4}}+\frac 1{\sqrt[3]{5}}+\cdots+\frac 1{\sqrt[3]{100}}

Find s \lfloor s \rfloor

Inspiration


The answer is 31.

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1 solution

Zakir Husain
May 31, 2020

I will give 2 2 solutions

First solution

It will be easy if you know calculus

s 1 1 100 x 1 / 3 d x s s-1≤∫_{1}^{100}x^{-1/3}dx≤s s 1 1 100 d 3 2 x 2 3 d x d x s s-1≤∫_{1}^{100}\frac{d\frac{3}{2}x^{\frac{2}{3}}}{\cancel{dx}}\cancel{dx}≤s s 1 1 100 d 3 2 x 2 3 s s-1≤∫_{1}^{100}d\frac{3}{2}x^{\frac{2}{3}}≤s s 1 3 2 10 0 2 3 2 1 2 = 3 2 ( 10 0 2 1 ) = 30.8165204... s s-1≤\frac{3}{2}\sqrt{100^2}-\frac{3}{2}\sqrt{1^2}=\frac{3}{2}(\sqrt{100^2}-1)=30.8165204...≤s s 31 s\approx31

Second solution

Through the coding environment in Brilliant:

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