Question

b) 1+2+3...+20
c) 1+5+9+...+101
d) 714+72+724
DAU CORONITA

Answer 1

$\boxed{1+2+3+...+n = \dfrac{n\cdot (n+1)}{2}}\rightarrow Suma ~lui~Gauss\\ \\ \\ b)\quad 1+2+3+...+20 =\dfrac{20\cdot (20+1)}{2} = 10\cdot(20+1) = 10\cdot 21 = 210\\ \\ c)\quad 1+5+9+...+101 \overset{(*)}{=} \\ \\( creste din 4 in 4, il vom avea pe 4 ca factor,iar scazatorul trebuie sa \\ il intuim) \\ \\ \overset{(*)}{=} (4\cdot 1-3)+(4\cdot 2-3) +(4\cdot 3-3)+...+(4\cdot 26-3) = \\ \\ = 4\cdot (1+2+3+...+26)-3-3-3-\underset{\text{de 26 ori}}{\underbrace{...}}-3 =$
$\\ \\ = 4\cdot \dfrac{26\cdot (26+1)}{2} - 3\cdot 26 = \\ \\ = 2\cdot 26\cdot (26+1)-3\cdot 26 = \\ \\ = 2\cdot 26\cdot 27-3\cdot 26= \\ \\ = 26\cdot (2\cdot 27-3) = \\ \\ = 26\cdot (54-3) = \\ \\ = 26\cdot 51 = \\ \\ = 1326$


$d)\quad 714+72+724 = 1510$