Question

Sa se demonstreze ca pentru orice x apartine lui R numerele 3^x-1,3^x+1 si 5*3^x+1sunt termeni consecutivi intr-o progresie aritmetica

Answer 1

$3^x+1= \frac{3^x-1+5*3^x+1}{2}$

Notam $3^x=t$

[tex]t+1= \frac{t-1+5t+1}{2} [/tex]

$t+1= \frac{6t}{2}$

[tex]t+1=3t t-3t+1=0 -2t+1=0 -2t=-1 /(-1) 2t=1 t=1/2[/tex]

[tex]3^x= \frac{1}{2} x=log_{3} \frac{1}{2}=log_{3}2-log_{3} 1= log_{3}2[/tex]


Si daca ai tot asa si se cere si progresie geometrica faci conform formulei

>>>>> $b^2=a*c$