tugas pak boko

matematika induksi bukti 2 pangkat n lebih dari n pangkat 3

Langkah: Apakah 2^(k+1) > (k+1)^3?
====> 2^(k+1) = 2.2^k > 2k^3
============ii2.2^k > k^3 + k^3
============ii2.2^k > k^3 + k.k^2
=====iKarena k>=10, maka:
============iik^3 + k.k^2 > k^3 + 4k^2 (bukan 4 juga ga papa, pokoknya bilangan dibawah 10)
============iik^3 + k.k^2 > k^3 + 3k^2 + k^2
============iik^3 + k.k^2 > k^3 + 3k^2 + k.k
=====iKarena k>=10 (lagi), maka:
============iik^3 + 3k^2 + k.k > k^3 + 3k^2 + 4k
============iik^3 + 3k^2 + k.k > k^3 + 3k^2 + 3k + k
=====iKarena k>=10 (lagi?), maka:
============iik^3 + 3k^2 + 3k + k > k^3 + 3k^2 + 3k + 1

Jadi, 2^(k+1) > (k+1)^3 terbukti sudah kebenarannya.

cara ke 2

hiupotesis: Anggap benar 2^k > k^3

Langkah: Apakah 2^(k+1) > (k+1)^3?
====> 2^(k+1) = 2.2^k > (k+1)^3
============ii2.k^3 > (k+1)^3 (substitusi dari hipotesis)
============iik^3 + k.k^2 > k^3 + 3k^2 + 3k + 1
=====iKarena k>=10, maka:
============iik^3 + 10k^2 > k^3 + 3k^2 + 3k + 1
============iik^3 + 3k^2 + 7k.k > k^3 + 3k^2 + 3k+ 1
=====iKarena k>=10 (lagi), maka:
============iik^3 + 3k^2 + 70k > k^3 + 3k^2 + 3k + 1
============iik^3 + 3k^2 + 3k + 67.k > k^3 + 3k^2 + 3k +1
=====iKarena k>=10 (lagi?), maka:
============iik^3 + 3k^2 + 3k + 670 > k^3 + 3k^2 + 3k + 1

Jadi, 2^(k+1) > (k+1)^3 terbukti sudah kebenarannya.

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Posted by hariez, Published at 20.19 and have 0 komentar