Question

Rezolvati ecuatia, unde x apartine N*
$1 + \frac{1}{1 + 2} + \frac{1}{1 + 2 + 3} + ... + \frac{1}{1 + 2 + 3 + ... + x} = \frac{200}{101}$
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Answer 1

$\displaystyle \sum\limits_{k=1}^{x}\dfrac{1}{\sum\limits_{i=1}^ki} = \dfrac{200}{101}\\ \\\\\Rightarrow \sum\limits_{k=1}^{x}\dfrac{1}{\dfrac{k(k+1)}{2}}= \dfrac{200}{101} \\ \\ \Rightarrow \sum\limits_{k=1}^{x}\dfrac{2}{k(k+1)} = \dfrac{200}{101}\\ \\\Rightarrow \sum\limits_{k=1}^{x}\dfrac{1}{k(k+1)}=\dfrac{100}{101}\\ \\ \Rightarrow \sum\limits_{k=1}^{x}\Big(\dfrac{1}{k}-\dfrac{1}{k+1}\Big)$

$\Rightarrow \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{2}-\dfrac{1}{3}-...-\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{100}{101}\\ \\ \Rightarrow 1-\dfrac{1}{x+1} = \dfrac{100}{101}\\ \\ \Rightarrow \dfrac{x+1-1}{x+1}=\dfrac{100}{101}\\ \\ \Rightarrow \dfrac{x}{x+1}=\dfrac{100}{101}\\ \\ \\ \Rightarrow \boxed{x = 100}$