Soal Induksi Matematika Disktrit, Buktikan bahawa : 3 .50+3.51+3.52+…+3.5n= 3(5 n+1-1) / 4  dimana n >= 0

1. Basis Induksi, dengan n =0 maka didapatkan

  • 3 .5= 3 akan sama dengan
  • 3(5 n+1-1) / 4 = 3. (50+1  – 1) / 4 = 12 / 4 = 3 ==> BENAR

2. Langkah Induksi, untuk n+1

  • 3 .50+3.51+3.52+…+3.5n= 3(5 n+1-1) / 4
  • 3 .50+3.51+3.52+…+3.5+ 3. 5n+1 = 3(5 n+2-1) / 4
  • 3(5 n+1-1) / 4 + 3. 5n+1 = 3(5 n+2-1) / 4
  • (3(5 n+1-1)  + 4. 3. 5n+1 ) / 4 = 3 (5 n+2 – 1) / 4
  • (3.5 n+1 – 3  + 12. 5n+1 ) / 4 = 3 (5 n+2 – 1) / 4
  • (3.5 n+1 + 12. 5n+1 – 3) / 4 = 3 (5 n+2 – 1) / 4
  • (15.5 n+1 – 3) / 4 = 3 (5 n+2 – 1) / 4   à (15 merupakan 3 x 5 atau 3 x 5 1)
  • (3.5 1.5 n+1 – 3) / 4 = 3 (5 n+2 – 1) / 4
  • (3.5 n+2 – 3) / 4 = 3 (5 n+2 – 1) / 4     à (5 1.5 n+1  sama dengan 5 n+1+1 )
  • 3 (5 n+2 – 1) / 4 = 3 (5 n+2 – 1) / 4 ==> terbukti BENAR

(Sebelah kirim sudah sama dengan kanan, jadi terbukti). Kedua langkah memberikan bukti benar, maka kesimpulan terbukti.

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