Wildan Bagus W 6 Report post Posted November 6, 2017 Diketahui \[S=1! \cdot 1 + 2! \cdot 2 + 3! \cdot 3 + ... + k! \cdot k\] dengan $k \in \mathbb{N}$. Buktikan bahwa sisa dari $\frac{S}{k+1}$ adalah $k$. 1 Share this post Link to post Share on other sites
yeftanma 0 Report post Posted November 7, 2017 Pake teorema wilson yaa? Tp kayak begimana yahh? Share this post Link to post Share on other sites
Wildan Bagus W 6 Report post Posted November 7, 2017 1 hour ago, yeftanma said: Pake teorema wilson yaa? Tp kayak begimana yahh? Bisa pakai teleskopik 1 Share this post Link to post Share on other sites
OnePunchMan 1 Report post Posted December 27, 2017 On 11/7/2017 at 2:56 PM, Wildan Bagus W said: Bisa pakai teleskopik Dari hint: Karena $m! \cdot m = (m+1)! - m!$ maka $S = (k+1)! -1$. Jadi $S \cong -1\mod (k+1)$. Akibatnya $S \cong k \mod k$. nulis garis 3 (kongruensi modulo) gimana sih? Share this post Link to post Share on other sites
Wildan Bagus W 6 Report post Posted December 28, 2017 8 hours ago, OnePunchMan said: Dari hint: Karena m!⋅m=(m+1)!−m!m!⋅m=(m+1)!−m! maka S=(k+1)!−1S=(k+1)!−1 . Jadi S≅−1mod(k+1)S≅−1mod(k+1) . Akibatnya S≅kmodkS≅kmodk . nulis garis 3 (kongruensi modulo) gimana sih? Itu pakai \equiv, jadnya $\equiv$ 1 Share this post Link to post Share on other sites
OnePunchMan 1 Report post Posted December 29, 2017 On 12/28/2017 at 7:00 AM, Wildan Bagus W said: Itu pakai \equiv, jadnya ≡ terima kasih. kirain pakai \cong, soalnya bacanya kongruen Share this post Link to post Share on other sites