Question

rog seriozitate va rogggg​

Answer 1

10.

în paranteză avem n-2+1 = n - 1 termeni

observăm că putem grupa termenii astfel:

$\dfrac{6}{8} - \dfrac{^{4)} 1}{2} = \dfrac{6 - 4}{8} = \dfrac{2^{(2}}{8} = \dfrac{1}{4}$

$\dfrac{7}{12} - \dfrac{^{4)} 1}{3} = \dfrac{7 - 4}{12} = \dfrac{3^{(3}}{12} = \dfrac{1}{4}$

...

$\dfrac{n + 4}{4n} - \dfrac{^{4)} 1}{n} = \dfrac{n + 4 - 4}{4n} = \dfrac{n^{(n} }{4n} = \dfrac{1}{4}$

$a = \dfrac{5}{4} + \dfrac{6}{8} + \dfrac{7}{12} + ... + \dfrac{n + 4}{4n} - \bigg(\dfrac{1}{2} + \dfrac{1}{3} + ... + \dfrac{1}{n} \bigg) =$

$= \dfrac{5}{4} + \bigg(\dfrac{6}{8} - \dfrac{1}{2}\bigg) + \bigg(\dfrac{7}{12} - \dfrac{1}{3}\bigg) + ... + \bigg(\dfrac{n + 4}{4n} - \dfrac{1}{n} \bigg)$

$= \dfrac{5}{4} + \underbrace{\dfrac{1}{4} + \dfrac{1}{4} + ... + \dfrac{1}{4}}_{n-1} = \dfrac{5}{4} + (n - 1) \cdot \dfrac{1}{4}$

$= \dfrac{5 + n - 1}{4} = \bf\dfrac{n + 4}{4}$

11. folosim formula:

$\boxed{\dfrac{1}{n \cdot (n + 1)} = \dfrac{1}{n} - \dfrac{1}{n + 1}}$

$\dfrac{1}{2} = \dfrac{1}{1 \cdot 2} = \dfrac{1}{1} - \dfrac{1}{2}$

$\dfrac{1}{6} = \dfrac{1}{2 \cdot 3} = \dfrac{1}{2} - \dfrac{1}{3}$

$\dfrac{1}{12} = \dfrac{1}{3 \cdot 4} = \dfrac{1}{3} - \dfrac{1}{4}$

...

$\dfrac{1}{9900} = \dfrac{1}{99 \cdot 100} = \dfrac{1}{99} - \dfrac{1}{100}$

$S = \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + \dfrac{1}{20} + ... + \dfrac{1}{9900}$

$= \dfrac{1}{1 \cdot 2} + \dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} + ... + \dfrac{1}{99 \cdot 100}$

$= \dfrac{1}{1} - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{99} - \dfrac{1}{100}$

se reduc termenii asemenea:

$= \dfrac{1}{1} - \dfrac{1}{100} = \dfrac{100 - 1}{100} = \bf \dfrac{99}{100}$