Question

Ajutor va rog la aceste problene. Va rog seriozitate. Probleme sunt 2 si 3 va rog​

Answer 1

Răspuns:

Explicație pas cu pas:

sume Gauss:

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

$\boxed{ 1 + 3+ 5 + ... + (2n - 1) = {n}^{2}}$

2)

$\dfrac{1}{100} + \dfrac{2}{100} + ... + \dfrac{99}{100} + \dfrac{1}{625} + \dfrac{3}{625}+ ... + \dfrac{49}{625} =$

$= \dfrac{1+2+...+99}{100} + \dfrac{1+3+...+49}{625} = \dfrac{\dfrac{99 \cdot 100}{2}}{100} + \dfrac{(2 \cdot 25 - 1)}{625}$

$= \dfrac{99 \cdot 100}{2 \cdot 100} + \dfrac{25^{2} }{625} = \dfrac{99}{2} + \dfrac{625}{625} = 49\dfrac{1}{2} + 1$

$=\bf 50\dfrac{1}{2}$

3)

$\dfrac{31}{10} + \dfrac{34}{11} + \dfrac{271}{90} - \bigg(\dfrac{1}{10} + \dfrac{1}{11}+ ... + \dfrac{1}{90}=\bigg)$

grupăm termenii:

$= \bigg(\dfrac{31}{10} - \dfrac{1}{10}\bigg) + \bigg(\dfrac{34}{11} - \dfrac{1}{11}\bigg) + ... + \bigg(\dfrac{271}{90} - \dfrac{1}{90}\bigg)$

$= \dfrac{31-1}{10} + \dfrac{34-1}{11} + ... + \dfrac{271-1}{90} = \dfrac{30}{10} + \dfrac{33}{11} + ... + \dfrac{270}{90}$

(stabilim numărul de termeni din sumă: 90-10+1=81)

$= \underbrace{3 + 3 + ... + 3}_{81} = 3 \cdot 81 =\bf 243$