Question

14. Dacă √(a−1)² + √(a₂-3)² +√(a₂-5)³² +...+ √(a1011-2021)² ≤0, atunci a₁ + a₂ +...+1
a1011 este:
a. număr par;
b. cub perfect;
c. pătrat perfect;
d. număr negativ.​

Answer 1

Răspuns:

c. pătrat perfect

Explicație pas cu pas:

$\sqrt{(a_1-1)^2}+\sqrt{(a_2-3)^2}+...+\sqrt{(a_{1011}-2021)^2} \leq 0\\|a_1-1|+|a_2-3|+...+|a_{1011}-2021|\leq 0\\\underbrace{|a_1-1|}_{ \geq 0} , ... , \underbrace{|a_{1011}-2021|}_{\geq 0} \implies |a_1-1|=0, ... , |a_{1011}-2011|=0\\a_1-1=0, a_1=1\\...\\a_{1011}-2021=0, a_{1011}=2021$

$a_1+a_2+...+a_{1011} = \underbrace{1+3+5+...+2021}_{1011 \ termeni}=\frac{(1+2021)\cdot1011}{2}=\frac{(a_1+a_n)\cdot n}{2} =1011\cdot1011=1011^2$

$n=\frac{2021-1}{2} +1=1011$