Question

să se calculeze radical din A, daca:

Answer 1

$A = \dfrac{1}{1\cdot4}+\dfrac{1}{2\cdot 6}+\dfrac{1}{3\cdot 8}+\dfrac{1}{4\cdot 10}+...+\dfrac{1}{48\cdot 98}+\dfrac{1}{49\cdot 100} \\ \\ A = \sum\limits_{k=1}^{49}\dfrac{1}{k\cdot (2k+2)} = \sum\limits_{k=1}^{49}\dfrac{1}{k\cdot 2(k+1)} = \sum\limits_{k=1}^{49}\dfrac{1}{2}\cdot\dfrac{1}{k\cdot (k+1)} =$
$=\dfrac{1}{2}\cdot \sum\limits_{k=1}^{49}\dfrac{1}{k\cdot (k+1)} = \dfrac{1}{2}\cdot \sum\limits_{k=1}^{49}\Big(\dfrac{1}{k}-\dfrac{1}{k+1}\Big)}= \\ \\ =\dfrac{1}{2}\cdot\Big(\sum\limits_{k=1}^{49}\dfrac{1}{k}-\sum\limits_{k=1}^{49}\dfrac{1}{k+1}\Big) = \\ \\ = \dfrac{1}{2}\cdot \Big(\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}-\Big(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{50}\Big)\Big) =$

$= \dfrac{1}{2}\cdot\Big(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}-\Big(\dfrac{1}{2}+\dfrac{1}{3}+ \dfrac{1}{4}+...+\dfrac{1}{49}\Big)-\dfrac{1}{50}\Big) = \\ \\ = \dfrac{1}{2}\cdot \Big(1-\dfrac{1}{50}\Big) = \dfrac{1}{2}\cdot \Big(\dfrac{50-1}{50} \Big)= \dfrac{1}{2}\cdot \dfrac{49}{50} = \dfrac{49}{100} \\ \\ \Rightarrow \sqrt{A} = \sqrt{\dfrac{49}{100}}= \dfrac{7}{10}$


[tex]\\ $M-am folosit de proprietatile sumei (Sigma): \\ \\ \left\| \begin{array}{c} \sum\limits_{\big{k=1}}^{\big{n}}a\cdot k =a\cdot \sum\limits_{\big{k=1}}^{\big{n}} k \quad\quad \quad \quad $ $ \\ \sum\limits_{\big{k=1}}^{\big{n}}\big(a+b\big) = \sum\limits_{\big{k=1}}^{\big{n}}a+\sum\limits_{\big{k=1}}^{\big{n}}b\end{array} \right |[/tex]


Cu suma (Sigma) se rezolva foarte lejer.

Answer 2

[tex]\it \dfrac{1}{1\cdot4} +\dfrac{1}{2\cdot6} +\dfrac{1}{3\cdot8} +\ ...\ \dfrac{1}{49\cdot100} = \\\;\\ \\\;\\ =\dfrac{1}{2} \left(\dfrac{1}{1\cdot2} +\dfrac{1}{2\cdot3} +\dfrac{1}{3\cdot4} +\ ...\ \dfrac{1}{49\cdot50} \right) [/tex]

Fiecare fracție din paranteză se descompune după formula:

$\it \dfrac{1}{k(k+1)} =\dfrac{1}{k}-\dfrac{1}{k+1}$

Prin urmare, vom avea:


[tex]\it \dfrac{1}{2} \left(\dfrac{1}{1}-\dfrac{1}{2} +\dfrac{1}{2}-\dfrac{1}{3} +\dfrac{1}{3} -\dfrac{1}{4}+\ ...\ \dfrac{1}{49} -\dfrac{1}{50} \right) =\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{50}\right)= \\\;\\ \\\;\\ =\dfrac{1}{2}\cdot\dfrac{49}{50} =\dfrac{49}{100}[/tex]

Se cere radical din acest ultim rezultat:

$\it \sqrt{\dfrac{49}{100}} =\dfrac{\sqrt{49}}{\sqrt{100}} =\dfrac{7}{10}.$