Question

rog seriozitate va rog mult 2 si 3​

Answer 1

2.observăm că putem grupa termenii:

$n = \bigg(\dfrac{12}{11} - ^{11)}1 \bigg) + \bigg(\dfrac{13}{22} - \dfrac{^{11)}1}{2} \bigg) + \bigg(\dfrac{14}{33} - \dfrac{^{11)}1}{3} \bigg) + ... + \bigg(\dfrac{110}{1089} - \dfrac{^{11)}1}{99} \bigg) =$

$= \bigg(\dfrac{12}{11} - \dfrac{11}{11} \bigg) + \bigg(\dfrac{13}{22} - \dfrac{11}{22} \bigg) + \bigg(\dfrac{14}{33} - \dfrac{11}{33} \bigg) + ... + \bigg(\dfrac{110}{1089} - \dfrac{11}{1089} \bigg)$

$= \dfrac{12 - 11}{11} + \dfrac{13 - 11}{22} + \dfrac{14 - 11}{33} + ... + \dfrac{110 - 11}{1089}$

$= \dfrac{1}{1 \cdot 11} + \dfrac{2}{2 \cdot 11} + \dfrac{3}{3 \cdot 11} + ... + \dfrac{99}{99 \cdot 11}$

$= \underbrace{\dfrac{1}{11} + \dfrac{1}{11} + \dfrac{1}{11} + ... + \dfrac{1}{11}}_{99} = 99 \cdot \dfrac{1}{11}$

$= \bf 9$

3.folosim formula sumei Gauss:

$\boxed {1 + 2 + 3 + ... + n = \dfrac{n \cdot (n + 1)}{2}}$

$a = \dfrac{1}{1 + 2} + \dfrac{1}{1 + 2 + 3} + \dfrac{1}{1 + 2 + 3 + 4} + ... + \dfrac{1}{1 + 2 + 3 + ... + 100} =$

$= \dfrac{1}{\dfrac{2 \cdot 3}{2}} + \dfrac{1}{\dfrac{3 \cdot 4}{2}} + \dfrac{1}{\dfrac{4 \cdot 5}{2}} + ... + \dfrac{1}{\dfrac{100 \cdot 101}{2}}$

$= \dfrac{2}{2 \cdot 3} + \dfrac{2}{3 \cdot 4} + \dfrac{2}{4 \cdot 5} + ... + \dfrac{2}{100 \cdot 101}$

$= 2 \cdot \bigg(\dfrac{1}{2 \cdot 3} + \dfrac{1}{3 \cdot 4} + \dfrac{1}{4 \cdot 5} + ... + \dfrac{1}{100 \cdot 101} \bigg)$

folosim formula:

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

$= 2 \cdot \bigg(\dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \dfrac{1}{4} - \dfrac{1}{5} + ... + \dfrac{1}{100} - \dfrac{1}{101} \bigg)$

se reduc termenii asemenea

$= 2 \cdot \bigg(\dfrac{1}{2} - \dfrac{1}{101} \bigg) = \dfrac{1}{1} - \dfrac{2}{101} = \dfrac{101 - 2}{101}$

$= \bf \dfrac{99}{101}$