Question

Sa se calculeze suma 3 + 5 + 7+ ... +21

Answer 1

$\boxed{\text{Metoda cu suma lui Gauss:} 1+2+3+...+n = \dfrac{n\cdot (n+1)}{2}}\\\\3 + 5 + 7+ ... +21 = \\ \\ = (1\cdot 2+1)+(2\cdot 2+1)+(2\cdot 3+1)+...+(2\cdot 10+1) = \\ \\ = 1\cdot 2+2\cdot 2+3\cdot 2+...+2\cdot 10+1+1+\underset{de ~10~ ori}{\underbrace{...}}+1 = \\ \\ = 2\cdot (1+2+3+...+10)+1\cdot 10 = \\ \\ = 2\cdot \dfrac{10\cdot (10+1)}{2}+10 = \\ \\ = 10\cdot (10+1)+10 = \\ \\ = 10\cdot 11+10 = \\ \\ = 10\cdot (11+1) = \\ \\ =10 \cdot 12 = \\ \\ = 120$


$\boxed{\text{Metoda cu formula: } 1+3+5+7+...+(2\cdot n-1) = n^2}} \\ \\ 3+5+7+...+21 = 1+3+5+7+...+21 - 1 = \\ \\ = 1+3+5+7+...+(2\cdot 11-1) - 1 = \\ \\ = 11^2 - 1 = \\ \\ = 121-1 = \\ \\ = 120$


$\boxed{\text{Metoda cu progresia aritmetica: } S_n = \dfrac{(a_1+a_n)\cdot n}{2}}}$

$S_n =3+5+7 +...+21 \\ \\ \left\| \begin{array}{ll} a_1 = 3 \\ r = 2 \\ a_n = 21 \\ \\ a_n = a_1+ (n-1)\cdot r \Rightarrow \\ \Rightarrow 21 = 3+(n-1)\cdot 2 \Rightarrow \\ \Rightarrow 21 = 3+2n-2 \Rightarrow \\ \Rightarrow 2n = 21-1 \Rightarrow \\ \Rightarrow n = 10\end{array} \right | \left\ \begin{array}{ll} S_{10} = \dfrac{(3+21)\cdot 10}{2} =\dfrac{24\cdot 10}{2} = \\ \\ =\dfrac{240}{2} = 120 \end{array} \right$

Answer 2

Se scrie suma
S= 3+5+7+...+21
S=21+19+17+3
————————— + le aduni
2S=24+24+24+...+24
Numarul de termeni este: (21-3) :2+1 (ultimul minus primul) : la rație +1
18:2+1=10
2S=24•10
2S=240 | :2
S=120