Cum se rezolva corect?

[tex]S_n = n^2+an+b\\ \\ a_1 = 2 \Rightarrow S_1 = 2 \Rightarrow 1+a+b =2 \Rightarrow a+b=1 \\ \\ S_2 = a_1+a_2 \Rightarrow 4+2a+b = 2+a_2 \Rightarrow a_2 = 2a+b+2 = a+3\\ \\ S_3 = a_1+a_2+a_3 \Rightarrow 9+3a+b = 2+a+3+a_3 \Rightarrow\\ \Rightarrow a_3 = 2a+b+4 = a+5 \\ \\ 2a_2 = a_1+a_3\Rightarrow 2a+6 = 2+a+5 \Rightarrow \boxed{a = 1} \\ \\ a+b = 1 \Rightarrow \boxed{b = 0}[/tex]

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Targoviste44 Targoviste44 · Answer 2 · 2019-05-05T02:21:46+03:00

I)

[tex]\it S_n=n^2+an+b\ \ \ \ (*)\\ \\ a_1=2\ \ \ \ (1)\\ \\ a_1=S_1=1+a+b\ \ \ \ (2)\\ \\ (1),(2) \Rightarrow 1+a+b=2 \Rightarrow a+b=1 \Rightarrow b=1-a\ \ \ \ \ (3)\\ \\ a_2=S_2-a_1=4+2a+b-2=2a+b+2\ \stackrel{(3)}{=}\ 2a+1-a+2=a+3\\ \\ a_3=S_3-S_2=9+3a+b-4-2a-b=a+5[/tex]

[tex]\it \div\ \ \dfrac{a_1+a_3}{2}=a_2 \Rightarrow \dfrac{2+a+5}{2}=a+3 \Rightarrow a+7=2a+6 \Rightarrow a=1\ \ \ (4)\\ \\ \\ (3),(4) \Rightarrow b=0[/tex]

Varianta corectă este c) a = 1, b = 0.

II)

[tex]\it a_1=2\ \ \ \ (1)\\ \\ a_1=S_1=1+a+b\ \ \ (2)\\ \\ (1),\ (2) \Rightarrow 1+a+b=2 \Rightarrow a+b=1\ \ \ \ (3)[/tex]

Singura variantă din grilă care verifică relația (3) este c).