Question

fie legea de compozitie
x*y = 2xy-3x-3y+6
aflati n pentru carr
(2^n +3/2)* (2^(n+1)+3/2)* (2^(n+2)+3/2) = 2^20+ 3/2

legea se poate scrie si ca
x*y = 2(x-3/2)(y-3/2)+3/2
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Answer 1

Răspuns:

Explicație pas cu pas:

Vom aplica x*y = 2(x-3/2)(y-3/2)+3/2

$(2^{n}+\frac{3}{2})*(2^{n+1}+\frac{3}{2})=2(2^{n}+\frac{3}{2}-\frac{3}{2})(2^{n+1}+\frac{3}{2}-\frac{3}{2})+\frac{3}{2}=2*2^{n}*2^{n+1}+\frac{3}{2}=2^{1+n+n+1}+ \frac{3}{2}=2^{2n+2}+ \frac{3}{2}.\\(2^{n}+\frac{3}{2})*(2^{n+1}+\frac{3}{2})*(2^{n+2}+\frac{3}{2})=(2^{2n+2}+ \frac{3}{2})*(2^{n+2}+ \frac{3}{2})=2(2^{2n+2}+ \frac{3}{2}-\frac{3}{2})(2^{n+2}+ \frac{3}{2}-\frac{3}{2})=2*2^{2n+2}*2^{n+2}+\frac{3}{2}=2^{1+2n+2+n+2}+\frac{3}{2}=2^{3n+5}+\frac{3}{2}\\Deci,~~2^{3n+5}+\frac{3}{2}=2^{20}+\frac{3}{2},~=>$

3n+5=20, ⇒3n=15, ⇒n=5.