Rješenja zadatka "Općinsko natjecanje 2000 SŠ3 4" korisnika "ceva18"

Neocijenjeno

16. prosinca 2023. 14:23 (4 mjeseci, 2 tjedna)

Korisnik: ceva18
Zadatak: Općinsko natjecanje 2000 SŠ3 4 (Sakrij tekst zadatka)

Dana je funkcija $f(x)=\displaystyle{\frac{a^x}{a^x+\sqrt{a}}}$ , gdje je $a$ pozitivan realan broj. Odredite $S=f\left(\frac{1}{2001}\right)+f\left(\frac{2}{2001}\right)+ \ldots + f\left(\frac{2000}{2001}\right).$

Upozorenje: Ovaj zadatak još niste riješili!
Kliknite ovdje kako biste prikazali rješenje.

Izračunajmo vrijednosti funkcija koje čine sumu: $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{a^{\frac{1999}{2001}}}{a^{\frac{1999}{2001}}+\sqrt{a}}+ \frac{a^{\frac{2000}{2001}}}{a^{\frac{2000}{2001}}+\sqrt{a}}$ Promotrimo li eksponente, gledajući "od nazad" sumu, vidjet ćemo da vrijedi: $\frac{2000}{2001} = 1 - \frac{1}{2001}$ , $\frac{1999}{2001} = 1 - \frac{2}{2001}$ itd. Uvrstimo: $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{a^{1-\frac{2}{2001}}}{a^{1-\frac{2}{2001}}+\sqrt{a}}+ \frac{a^{1-\frac{1}{2001}}}{a^{1-\frac{1}{2001}}+\sqrt{a}}$ $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{a^{1}\cdot a^{-\frac{2}{2001}}}{a^{1}\cdot a^{-\frac{2}{2001}} + \sqrt{a}} + \frac{a^{1}\cdot a^{-\frac{1}{2001}}}{a^{1}\cdot a^{-\frac{1}{2001}} + \sqrt{a}}$ $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\frac{a}{a^{\frac{2}{2001}}}}{\frac{a}{a^{\frac{2}{2001}}} + \sqrt{a}}+\frac{\frac{a}{a^{\frac{1}{2001}}}}{\frac{a}{a^{\frac{1}{2001}}} + \sqrt{a}}$ Valja se prisjetiti da se može zapisati $a = \sqrt{a} \cdot \sqrt{a}$ , a onda primijetiti da se razlomci (gledajući "od nazad") mogu podijeliti s $\sqrt{a}$ ( $a>0$ prema uvjetu). $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\frac{\sqrt{a}}{a^{\frac{2}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{2}{2001}}} + 1}+\frac{\frac{\sqrt{a}}{a^{\frac{1}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{1}{2001}}} + 1}$ $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\frac{\sqrt{a}}{a^{\frac{2}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{2}{2001}}} + 1}+\frac{\frac{\sqrt{a}}{a^{\frac{1}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{1}{2001}}} + 1}$ Sredimo zadnje članove sume (i analogne ostale koji se također mogu srediti): $\frac{\frac{\sqrt{a}}{a^{\frac{1}{2001}}}}{\frac{\sqrt{a}+a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}}} = \frac{\sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}}$ . Suma poprima oblik: $S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\sqrt{a}}{a^{\frac{2}{2001}}+\sqrt{a}}+ \frac{\sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}}.$ Grupirajmo prvi i zadnji, drugi i predzadnji član, ... i dobijemo: $S = (\frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} +\frac{\sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}}) + (\frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}}+ \frac{\sqrt{a}}{a^{\frac{2}{2001}}+\sqrt{a}}) + ...$ $S = \frac{a^{\frac{1}{2001}}+ \sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}+\sqrt{a}}{a^{\frac{2}{2001}}+\sqrt{a}} + ...$ $S = 1 + 1 + ...$ Imamo 2000 članova, ali svaki par u zbroju daje jedan što znači da ukupno imamo 1000 jedinica, tj. suma izraza jednaka je: $S = 1000$ .

Izračunajmo vrijednosti funkcija koje čine sumu: $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{a^{\frac{1999}{2001}}}{a^{\frac{1999}{2001}}+\sqrt{a}}+ \frac{a^{\frac{2000}{2001}}}{a^{\frac{2000}{2001}}+\sqrt{a}}$$ Promotrimo li eksponente, gledajući "od nazad" sumu, vidjet ćemo da vrijedi: $\frac{2000}{2001} = 1 - \frac{1}{2001}$, $\frac{1999}{2001} = 1 - \frac{2}{2001}$ itd. Uvrstimo: $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{a^{1-\frac{2}{2001}}}{a^{1-\frac{2}{2001}}+\sqrt{a}}+ \frac{a^{1-\frac{1}{2001}}}{a^{1-\frac{1}{2001}}+\sqrt{a}}$$ $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{a^{1}\cdot a^{-\frac{2}{2001}}}{a^{1}\cdot a^{-\frac{2}{2001}} + \sqrt{a}} + \frac{a^{1}\cdot a^{-\frac{1}{2001}}}{a^{1}\cdot a^{-\frac{1}{2001}} + \sqrt{a}}$$ $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\frac{a}{a^{\frac{2}{2001}}}}{\frac{a}{a^{\frac{2}{2001}}} + \sqrt{a}}+\frac{\frac{a}{a^{\frac{1}{2001}}}}{\frac{a}{a^{\frac{1}{2001}}} + \sqrt{a}}$$ Valja se prisjetiti da se može zapisati $a = \sqrt{a} \cdot \sqrt{a}$, a onda primijetiti da se razlomci (gledajući "od nazad") mogu podijeliti s $\sqrt{a}$ ($a>0$ prema uvjetu). $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\frac{\sqrt{a}}{a^{\frac{2}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{2}{2001}}} + 1}+\frac{\frac{\sqrt{a}}{a^{\frac{1}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{1}{2001}}} + 1}$$ $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\frac{\sqrt{a}}{a^{\frac{2}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{2}{2001}}} + 1}+\frac{\frac{\sqrt{a}}{a^{\frac{1}{2001}}}}{\frac{\sqrt{a}}{a^{\frac{1}{2001}}} + 1}$$ Sredimo zadnje članove sume (i analogne ostale koji se također mogu srediti): $\frac{\frac{\sqrt{a}}{a^{\frac{1}{2001}}}}{\frac{\sqrt{a}+a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}}} = \frac{\sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}}$. Suma poprima oblik: $$S = \frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}} + ... + \frac{\sqrt{a}}{a^{\frac{2}{2001}}+\sqrt{a}}+ \frac{\sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}}.$$ Grupirajmo prvi i zadnji, drugi i predzadnji član, ... i dobijemo: $$S = (\frac{a^{\frac{1}{2001}}}{a^{\frac{1}{2001}}+\sqrt{a}} +\frac{\sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}}) + (\frac{a^{\frac{2}{2001}}}{a^{\frac{2}{2001}}+\sqrt{a}}+ \frac{\sqrt{a}}{a^{\frac{2}{2001}}+\sqrt{a}}) + ...$$ $$S = \frac{a^{\frac{1}{2001}}+ \sqrt{a}}{a^{\frac{1}{2001}}+\sqrt{a}} + \frac{a^{\frac{2}{2001}}+\sqrt{a}}{a^{\frac{2}{2001}}+\sqrt{a}} + ...$$ $$S = 1 + 1 + ... $$ Imamo 2000 članova, ali svaki par u zbroju daje jedan što znači da ukupno imamo 1000 jedinica, tj. suma izraza jednaka je: $S = 1000$.

Školjka