Točno
2. travnja 2016. 19:41 (8 godine, 3 mjeseci)
Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.
Kliknite ovdje kako biste prikazali rješenje.
Ako tvrdnja vrijedi tada postoji prirodni broj
takav da
![n\cdot 2^{n+1}+1 = x^2 \implies n\cdot2^n = (x-1)(x+1)](/media/m/b/c/f/bcfed2b75b172f54e994ecf7004d7cc5.png)
je neparan, a
i
su udaljeni točno za dva pa je jedan djeljiv sa
, a drugi sa
pa razlikujemo dva slučaja:
1)
,
gdje su
, ![ab = n](/media/m/c/5/8/c58272696662f4a2772bf302bce58c4d.png)
![x+1-(x-1) = b\cdot 2^n - 2a](/media/m/a/b/3/ab3368503d5ab4e5962f82ee342b6149.png)
![2 = b\cdot 2^n - 2a](/media/m/f/6/7/f6793407d48898d8c75ae61b5aa660d9.png)
Gledamo minimalnu vrijednost funkcije![b\cdot 2^n - 2a](/media/m/8/8/b/88ba0cc2e43eea4b66bc6376555d5ea5.png)
za
imamo minimalnu vrijednost ![b\cdot 2^n - 2a = 1*2 - 2 = 0](/media/m/b/7/e/b7e3d59414d569cfad9a83a3c1a90ffa.png)
za
imamo ![b\cdot 2^n - 2a = 1\cdot 2^2 - 2\cdot 2 = 0](/media/m/6/7/1/6713118974e398cf82aa4f8050f64e84.png)
za
imamo ![b\cdot 2^n - 2a = 1\cdot 2^3 - 2\cdot 3 = 2](/media/m/9/4/8/9488a0f55519f6b5c71ed00ce0cc1745.png)
za
imamo ![b\cdot 2^n - 2a = 1\cdot 2^4 - 2\cdot 4 = 8 > 2](/media/m/e/5/7/e57ac052d0beeab650b69adc9a2be618.png)
Za veće
nemoramo provjeravati jer
raste eksponencijalno, a
linearno pa se minimalna vrijednost povećava
Zaključujemo:![n\leq 3](/media/m/4/c/e/4ce4b58846c3538f60b5b6485815ce3a.png)
2)
,
gdje su
, ![ab = n](/media/m/c/5/8/c58272696662f4a2772bf302bce58c4d.png)
![x+1 - (x-1) = 2a - b\cdot 2^n](/media/m/e/0/a/e0a180d524d2035ea6479d238cc3099e.png)
![2 = 2a - b\cdot 2^n](/media/m/5/7/e/57e474dc1efa488829d58287bbe5f0e3.png)
Gledamo maksimalnu vrijednost funkcije![2a - b\cdot 2^n](/media/m/f/d/5/fd5b4b4d78dfd6aa51b581401a121417.png)
za
imamo ![2a - b\cdot 2^n = 2\cdot 4 - 1\cdot 2^4 = -8 < 2](/media/m/2/4/d/24d1380ad3b3aee33f175ee5f47ade15.png)
Za veće
nemoramo provjeravati jer
raste eksponencijalno, a
linearno pa se maksimalna vrijednost smanjuje
Zaključujemo:![n \leq 3](/media/m/7/3/c/73cd40bd8baa439603cae0ce159361a8.png)
Uvrštavajući sve
u
dobijemo da je
jedino rješenje.
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
![n\cdot 2^{n+1}+1 = x^2 \implies n\cdot2^n = (x-1)(x+1)](/media/m/b/c/f/bcfed2b75b172f54e994ecf7004d7cc5.png)
![x](/media/m/f/1/8/f185adeed9bd346bc960bca0147d7aae.png)
![x-1](/media/m/9/1/6/916131c67619164bacdcb2089bba712a.png)
![x+1](/media/m/d/e/0/de098a179aba20a11154296253a54cc2.png)
![2](/media/m/e/e/e/eeef773d19a3b3f7bdf4c64f501e0291.png)
![2^n](/media/m/8/e/a/8ea40429bb1e68f68f9e7a97fd5351f7.png)
1)
![x-1 = 2a](/media/m/6/4/b/64b485c2a94ab9d8418f197c7d75155c.png)
![x+1 = b \cdot 2^n](/media/m/1/4/b/14bd0af76d6e1b99c8941d84d61c43a0.png)
![a,b \in N](/media/m/6/9/1/691b6594ccdd8e6d1bcff57c1c94f68f.png)
![ab = n](/media/m/c/5/8/c58272696662f4a2772bf302bce58c4d.png)
![x+1-(x-1) = b\cdot 2^n - 2a](/media/m/a/b/3/ab3368503d5ab4e5962f82ee342b6149.png)
![2 = b\cdot 2^n - 2a](/media/m/f/6/7/f6793407d48898d8c75ae61b5aa660d9.png)
Gledamo minimalnu vrijednost funkcije
![b\cdot 2^n - 2a](/media/m/8/8/b/88ba0cc2e43eea4b66bc6376555d5ea5.png)
za
![n=1](/media/m/4/e/4/4e466fe58c2a8f6389234c5c673f069c.png)
![b\cdot 2^n - 2a = 1*2 - 2 = 0](/media/m/b/7/e/b7e3d59414d569cfad9a83a3c1a90ffa.png)
za
![n=2](/media/m/e/8/2/e82e9f27692b705eb8b201ab32bd0a83.png)
![b\cdot 2^n - 2a = 1\cdot 2^2 - 2\cdot 2 = 0](/media/m/6/7/1/6713118974e398cf82aa4f8050f64e84.png)
za
![n=3](/media/m/6/e/8/6e8cc663572ec564892ed13a28debcb1.png)
![b\cdot 2^n - 2a = 1\cdot 2^3 - 2\cdot 3 = 2](/media/m/9/4/8/9488a0f55519f6b5c71ed00ce0cc1745.png)
za
![n=4](/media/m/b/c/a/bca400804be45e04111a2997eb638b64.png)
![b\cdot 2^n - 2a = 1\cdot 2^4 - 2\cdot 4 = 8 > 2](/media/m/e/5/7/e57ac052d0beeab650b69adc9a2be618.png)
Za veće
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![b\cdot 2^n](/media/m/e/4/e/e4e773fee95781e931e162a71efc5d5e.png)
![2a](/media/m/1/d/f/1df3cab05590beeed7eb5242b804b599.png)
Zaključujemo:
![n\leq 3](/media/m/4/c/e/4ce4b58846c3538f60b5b6485815ce3a.png)
2)
![x-1 = b\cdot 2^n](/media/m/8/c/a/8caf196b30129104eaf147ac2c2880ca.png)
![x+1 = 2a](/media/m/f/8/5/f8513d2cb175669eae7c54cfa9663bab.png)
![a,b \in N](/media/m/6/9/1/691b6594ccdd8e6d1bcff57c1c94f68f.png)
![ab = n](/media/m/c/5/8/c58272696662f4a2772bf302bce58c4d.png)
![x+1 - (x-1) = 2a - b\cdot 2^n](/media/m/e/0/a/e0a180d524d2035ea6479d238cc3099e.png)
![2 = 2a - b\cdot 2^n](/media/m/5/7/e/57e474dc1efa488829d58287bbe5f0e3.png)
Gledamo maksimalnu vrijednost funkcije
![2a - b\cdot 2^n](/media/m/f/d/5/fd5b4b4d78dfd6aa51b581401a121417.png)
za
![n = 4](/media/m/a/8/5/a855b286759b88abcea3d3dbbfe88d81.png)
![2a - b\cdot 2^n = 2\cdot 4 - 1\cdot 2^4 = -8 < 2](/media/m/2/4/d/24d1380ad3b3aee33f175ee5f47ade15.png)
Za veće
![n](/media/m/a/e/5/ae594d7d1e46f4b979494cf8a815232b.png)
![b\cdot 2^n](/media/m/e/4/e/e4e773fee95781e931e162a71efc5d5e.png)
![2a](/media/m/1/d/f/1df3cab05590beeed7eb5242b804b599.png)
Zaključujemo:
![n \leq 3](/media/m/7/3/c/73cd40bd8baa439603cae0ce159361a8.png)
Uvrštavajući sve
![n \leq 3](/media/m/7/3/c/73cd40bd8baa439603cae0ce159361a8.png)
![n\cdot 2^{n+1}+1 = x^2](/media/m/1/9/e/19ed046e20b42c71591ec56a2e2fcc80.png)
![n=3](/media/m/6/e/8/6e8cc663572ec564892ed13a28debcb1.png)
Ocjene: (1)
Komentari:
grga, 18. travnja 2016. 19:31