Question

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2
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Answer 1

Răspuns:

Explicație pas cu pas:

Answer 2

$\displaystyle \lg\dfrac{2^2}{1\cdot 3}+\lg\dfrac{3^2}{2\cdot 4}+\lg\dfrac{4^2}{3\cdot 5}+...\lg\dfrac{99^2}{98\cdot 100}=\\ \\ =\sum\limits_{k=1}^{98}\lg\dfrac{(k+1)^2}{k\cdot (k+2)} \\ \\ =\lg\prod\limits_{k=1}^{98}\dfrac{(k+1)^2}{k\cdot (k+2)}\\ \\ =\lg\left(\dfrac{99}{1}\cdot \prod\limits_{k=1}^{98}\dfrac{(k+1)^2}{(k+1)\cdot (k+2)} \right)\\ \\ = \lg\left(\dfrac{99}{1}\cdot \dfrac{2}{100}\cdot \prod\limits_{k=1}^{98}\dfrac{(k+1)^2}{(k+1)\cdot (k+1)} \right)$

$\displaystyle =\lg\left(\dfrac{99\cdot 2}{100}\cdot\prod\limits_{k=1}^{98}1\right)\\ \\ =\lg\dfrac{99}{50}$