n un numar natural care da restul 12 prin impartire la 100
n=100k+12
Avem:
[tex](a+b)^2=M_a+b^n[/tex]
n²+4n+8=(100k+12)²+4(100k+12)+8
n²+4n+8=M₁₀₀+12²+4(M₁₀₀+12)+8
n²+4n+8=M₁₀₀+144+M₁₀₀+48+8
n²+4n+8=M₁₀₀+200
200:100=2 rest 0⇒ restul impartirii n²+4n+8 la 100 este 0
Alte exercitii gasesti aici: https://brainly.ro/tema/1051790
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