Question

Numerele b1, b2, ... bn sunt in progresie geometrica. Cate numere sunt daca :
b7 - b5 = 48
b5 + b6 = 48
Sn = 1023
iar numerele de mai sus sunt in sistem.

Answer 1

Răspuns:

10

Explicație pas cu pas:

$\left \{ {{b_{7}-b_{5}=48} \atop {b_{5}+b_{6}=48}} \right. ~\left \{ {{b_{5}*q^{2}-b_{5}=48} \atop {b_{5}+b_{5}*q}=48} \right. ~\left \{ {{b_{5}(q^{2}-1)=48} \atop {b_{5}(1+q)=48}} \right. ~\left \{ {{b_{5}(q-1)(q+1)=48} \atop {b_{5}(1+q)=48}} \right. ~48*(q-1)=48,~q-1=1,~q=2.\\Atunci~din~b_{5}(1+q)=48,~obtinem~b_{5}(1+2)=48,~b_{5}*3=48,~b_{5}=16.\\b_{5}=b_{1}*q^{4},~deci~b_{1}*2^{4}=16,~b_{1}*16=16,~b_{1}=1.~\\S_{n}=b_{1}*\frac{q^{n}-1}{q-1},~deci~1*\frac{2^{n}-1}{2-1} =1023,~2^{n}-1=1023,~2^{n}=1024,$

Deci n=10, deoarece 2¹⁰=1024.