Wykaz, ze reszta z dzielenia sumy kwadratow trzech kolejnych liczb naturalnych przez 3 jest rowna 2.
a² + (a + 1)² + (a + 2)² = a² + a² + 1 + 2a + a² + 4 + 4a = 3a² + 5 + 6a = 3(a² + a + 5/3)
3(a² + a + 5/3) : 3 = a² + a + 5/3 = a² + a + 1 r. 2
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n²+(n+1)²+(n+2)² = n² + n² + 2n + 1 + n² + 4n + 4 =
= 3n² + 6n + 3 + 2 = 3(n²+2n+1) + 2
niech m = n²+2n+1
(3m+2)/3 = m r 2
a² + (a + 1)² + (a + 2)² = a² + a² + 1 + 2a + a² + 4 + 4a = 3a² + 5 + 6a = 3(a² + a + 5/3)
3(a² + a + 5/3) : 3 = a² + a + 5/3 = a² + a + 1 r. 2