[Purdue POW] Buktikan bahwa bilangan bulat terdekat dengan 
Solusi
Karena
maka
Suku pertama jelas merupakan bilangan bulat, sedangkan suku kedua
Maka bilangan bulat terdekat dengan
Bilangan ini tidak habis dibagi
![n!\displaystyle\sum^{n}_{k=0}\dfrac{(-1)^k}{k!}=n\left[(n-1)!\displaystyle\sum^{n-1}_{k=0}\dfrac{(-1)^k}{k!}\right]+(-1)^n](http://s2.wordpress.com/latex.php?latex=n%21%5Cdisplaystyle%5Csum%5E%7Bn%7D_%7Bk%3D0%7D%5Cdfrac%7B%28-1%29%5Ek%7D%7Bk%21%7D%3Dn%5Cleft%5B%28n-1%29%21%5Cdisplaystyle%5Csum%5E%7Bn-1%7D_%7Bk%3D0%7D%5Cdfrac%7B%28-1%29%5Ek%7D%7Bk%21%7D%5Cright%5D%2B%28-1%29%5En&bg=ffffff&fg=000000&s=0 "n!\displaystyle\sum^{n}_{k=0}\dfrac{(-1)^k}{k!}=n\left[(n-1)!\displaystyle\sum^{n-1}_{k=0}\dfrac{(-1)^k}{k!}\right]+(-1)^n")
Bilangan ini habis dibagi
![n!\displaystyle\sum^{n}_{k=0}\dfrac{(-1)^k}{k!}=n(n-1)\left[(n-2)!\displaystyle\sum^{n-2}_{k=0}\dfrac{(-1)^k}{k!}\right]+(-1)^{n-1}\cdot n+(-1)^n=\displaystyle(n-1)\left\{n\left[(n-2)!\displaystyle\sum^{n-2}_{k=0}\dfrac{(-1)^k}{k!}\right]+(-1)^{n-1}\right\}](http://s1.wordpress.com/latex.php?latex=n%21%5Cdisplaystyle%5Csum%5E%7Bn%7D_%7Bk%3D0%7D%5Cdfrac%7B%28-1%29%5Ek%7D%7Bk%21%7D%3Dn%28n-1%29%5Cleft%5B%28n-2%29%21%5Cdisplaystyle%5Csum%5E%7Bn-2%7D_%7Bk%3D0%7D%5Cdfrac%7B%28-1%29%5Ek%7D%7Bk%21%7D%5Cright%5D%2B%28-1%29%5E%7Bn-1%7D%5Ccdot+n%2B%28-1%29%5En%3D%5Cdisplaystyle%28n-1%29%5Cleft%5C%7Bn%5Cleft%5B%28n-2%29%21%5Cdisplaystyle%5Csum%5E%7Bn-2%7D_%7Bk%3D0%7D%5Cdfrac%7B%28-1%29%5Ek%7D%7Bk%21%7D%5Cright%5D%2B%28-1%29%5E%7Bn-1%7D%5Cright%5C%7D&bg=ffffff&fg=000000&s=0 "n!\displaystyle\sum^{n}_{k=0}\dfrac{(-1)^k}{k!}=n(n-1)\left[(n-2)!\displaystyle\sum^{n-2}_{k=0}\dfrac{(-1)^k}{k!}\right]+(-1)^{n-1}\cdot n+(-1)^n=\displaystyle(n-1)\left\{n\left[(n-2)!\displaystyle\sum^{n-2}_{k=0}\dfrac{(-1)^k}{k!}\right]+(-1)^{n-1}\right\}")
bah….
Komentar oleh cs_1912 — 4 Maret 2008 @ 17.28