Cm : n^n-n^2+n-1 chia hết cho (n-1)^2 vs mọi n và n>1 05/09/2021 Bởi Reese Cm : n^n-n^2+n-1 chia hết cho (n-1)^2 vs mọi n và n>1
Ta co nⁿ – n² + n – 1 = nⁿ – n – (n² – 2n + 1) = nⁿ – n – (n – 1)² => ta can chung minh nⁿ – n chia het cho (n – 1)² Ta co nⁿ – n = n.[ n^(n – 1) – 1 ] = n.(n – 1).[ n^(n – 2) + n^(n – 3) + … + n + 1 ] Ta co n^(n – 2) + n^(n – 3) + … + n + 1 dong du voi 1 + 1 + … + 1 (n – 1 chu so 1) (mod n – 1) => n^(n – 2) + n^(n – 3) + … + n + 1 dong du voi n – 1 (mod n-1) => n^(n – 2) + n^(n – 3) + … + n + 1 = k(n – 1) => n.(n – 1).[ n^(n – 2) + n^(n – 3) + … + n + 1 ] = kn(n – 1)² => nⁿ – n = kn(n – 1)² => nⁿ – n chia het cho (n – 1)² Vậy (dpcm) Chúc you hok tốt ???????????????? Bình luận
Ta co nⁿ – n² + n – 1 = nⁿ – n – (n² – 2n + 1)
= nⁿ – n – (n – 1)²
=> ta can chung minh nⁿ – n chia het cho (n – 1)²
Ta co nⁿ – n = n.[ n^(n – 1) – 1 ] = n.(n – 1).[ n^(n – 2) + n^(n – 3) + … + n + 1 ]
Ta co n^(n – 2) + n^(n – 3) + … + n + 1 dong du voi 1 + 1 + … + 1 (n – 1 chu so 1) (mod n – 1)
=> n^(n – 2) + n^(n – 3) + … + n + 1 dong du voi n – 1 (mod n-1)
=> n^(n – 2) + n^(n – 3) + … + n + 1 = k(n – 1)
=> n.(n – 1).[ n^(n – 2) + n^(n – 3) + … + n + 1 ] = kn(n – 1)²
=> nⁿ – n = kn(n – 1)² => nⁿ – n chia het cho (n – 1)²
Vậy (dpcm)
Chúc you hok tốt ????????????????