Question

2+3-4+5+6-7+8+9-10+...+191+192-193​

Answer 1

 

2 + 3 - 4 + 5 + 6 - 7 + 8 + 9 - 10 + ... + 191 + 192 - 193 =

(Facem grupe de cate 3 numere)

= (2 + 3 - 4) + (5 + 6 - 7) + (8 + 9 - 10) + ... + (191 + 192 - 193) =

= 1 + 4 + 7 + ... + 190 = ?

$\displaystyle\bf\\\text{\bf~Calculam numarul de termeni:}\\\\n=\frac{190-1}{3}+1=\frac{189}{3}+1=63+1=64~\text{\bf~de termeni}\\\\\text{\bf~Aplicam Gauss pentru a calcula suma:}\\\\S=\frac{n(190+1)}{2}=\frac{64(190+1)}{2}=\frac{64\times191}{2}=32\times191=\boxed{\bf6112}$

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Answer 2

Metoda 1)

$2+3-4+5+6-7+8+9-10+...+191+192-193= \\ \\ =3+6+9+...+192 +(2-4)+(5-7)+(8-10)+...+(191-193)=\\ \\ =3(1+2+3+...+64)-2-2-2-\underset{de~64~ori}{\underbrace{...}}-2 = \\ \\ = 3\cdot \dfrac{64\cdot (64+1)}{2}-2\cdot 64 = \\ \\ =3\cdot 32\cdot 65-128 = \\ \\ = \boxed{6112}$

Metoda 2)

$S = 2+3-4+5+6-7+8+9-10+...+191+192-193\\ \\ \text{Consideram sirul }2,5,8,... \text{ este de forma 3n-1}\\\\\displaystyle S =\sum\limits_{k=1}^{n}\Big[(3k-1)+(3k)-(3k+1)\Big] \\ \\ 3n+1 = 193 \Rightarrow 3n = 192 \Rightarrow n = 64\\ \\ S = \sum\limits_{k=1}^{64}\Big[(3k-1)+(3k)-(3k+1)\Big] \\ \\ S = \sum\limits_{k=1}^{64}(3k-2)=3\sum\limits_{k=1}^{64}k - 2\cdot 64 = \\ \\ =3\cdot \dfrac{64\cdot 65}{2}-128 = 3\cdot 32\cdot 65 -128 = \boxed{6112}$