cum aflam ratia intr-o progresie aritmetica

Sa mai rezolvam niste probleme pentru dragii nostrii vizitatori.

1. David are un numar de jucarii. Triplul jumatatii acestui numar micsorat cu jumatatea jumatatii numarului respectiv devine 20. Cate jucarii are David?

Rezolvarea problemei:

Notam numarul jucariilor cu x
Si formam ecuatia: $3\cdot x\cdot\frac{1}{2}-\frac{1}{2}\cdot \frac{1}{2}\cdot x=20$
Asadar avem ecuatia: $\frac{3x}{2}-\frac{x}{4}=20|\cdot 4\Rightarrow \frac{3x}{2}\cdot 4-\frac{x}{4}\cdot 4=20\cdot 4\Rightarrow 3x\cdot 2-x=80\Rightarrow 6x-x=80\Rightarrow 5x=80\Rightarrow x=80:5\Rightarrow x=16$
Deci numarul jucariilor lui David este 16.

Acum efectuam proba: $\frac{3x}{2}=\frac{3\cdot 16}{2}=\frac{48}{2}=48:2=24$ triplul jumatatii acestui numar
Micsorat cu jumatatea jumatatii numarului respectiv $24-\frac{1}{4}\cdot 16=24-\frac{16}{4}=24-16:4=24-4=20$ devine 20, ceea ce se verifica.

2. S = 8+11+14+…+44
Observam ca termenii sumei se afla in progresie aritmetica.
Ca sa calculam suma de mai sus folosim progresiile aritmetice:
Astfel avem ca: $a_{1}=8, a_{2}=11.... a_{n}=44$
Mai intai aflam ratia, astfel avem : $r=a_{2}-a_{1}=11-8=3$
Dar cu formula termenului general stim ca $a_{n}=a_{1}+\left(n-1\right)\cdot r\Rightarrow 44=8+\left(n-1\right)\cdot 3\Rightarrow \left(n-1\right)\cdot 3=44-8\Rightarrow \left(n-1\right)\cdot 3=36\Rightarrow n-1=36:3\Rightarrow n-1=12\Rightarrow n=12+1\Rightarrow n=13$

Deci stim ca in suma avem 13 termenii, iar in progresie aritmetica suma primilor n termenii este $S_{n}=\frac{\left(a_{1}+a_{n}\right)\cdot n}{2}$
Dar noi cum avem 13 termenii, obtinem $S_{13}=\frac{\left(a_{1}+a_{13}\right)\cdot 13}{2}=\frac{\left(8+44\right)\cdot 13}{2}=\frac{52\cdot 13}{2}=\frac{676}{2}=338$

Asadar suma primilor 13 termenii este 338.
Deci suma S=8+11+14+…+44=338

Prezentam subiecte posibile simulare Bacalaureat 2014 pentru subiectul III
Se considera functia $f:\left(0, +\infty\right)\rightarrow R$ definita prin $f\left(x\right)=\frac{1}{x^{2}}+\frac{1}{\left(x+1\right)^{2}}$
a) Sa se calculeze $f^{'}\left(x\right), x\in\left(0,\infty\right)$
b) Sa se demonstreze ca functia f este descrescatoare pe intervalul $\left(0,+\infty\right)$
c) Sa se calculeze $\lim\limits_{x\to +\infty}{x^{3}f^{'}\left(x\right)}$
Solutie:
a) $f^{'}\left(x\right)=\frac{-1\cdot\left(x^{2}\right)^{'}}{\left(x^{2}\right)^{2}}+\frac{-1\cdot \left(\left(x+1\right)^{2}\right)^{'}}{\left[\left(x+1\right)^{2}\right]^{2}}=\frac{-1\cdot 2x}{x^{4}}+\frac{-1\cdot 2\left(x+1\right)\cdot \left(x+1\right)^{'}}{\left(x+1\right)^{4}}=\frac{-2x}{x^{4}}+\frac{-2\left(x+1\right)\cdot 1}{\left(x+1\right)^{4}}=\frac{-2}{x^{3}}-\frac{2}{\left(x+1\right)^{3}}=\frac{-2\left(x+1\right)^{3}-2\cdot x^{3}}{x^{3}\left(x+1\right)^{3}}=\frac{-2\left[\left(x+1\right)^{3}+x^{3}\right]}{x^{3}\left(x+1\right)^{3}}$ .
b) Observam ca $f^{'}\left(x\right)<0$ , deci f este strict descrescatoare pe $\left(0,+\infty\right)$
c) Acum sa calculam limita
$\lim\limits_{x\to +\infty}{x^{3}\cdot \left(\frac{-2\left[\left(x+1\right)^{3}+x^{3}\right]}{x^{3}\left(x+1\right)^{3}}\right)}=$
$\lim\limits_{x\to +\infty}{\frac{-2\left[\left(x+1\right)^{3}+x^{3}\right]}{\left(x+1\right)^{3}}}=$
$\lim\limits_{x\to +\infty}{-2\left[\cdot \frac{\left(x+1\right)^{3}}{\left(x+1\right)^{3}}+\frac{x^{3}}{\left(x+1\right)^{3}}\right]}=$
$-2\lim\limits_{x\to +\infty}{1+\frac{x^{3}}{\left(x+1\right)^{3}}}=-2\left(1+1\right)=-2\cdot 2=-4$ .
2) Se considera functia $f:\left(0, +\infty\right)\rightarrow R, f\left(x\right)=\frac{\ln x}{x}+x$
a) Sa se calculeze $\int^{e}_{1}\left(f\left(x\right)-\frac{\ln x}{x}\right)dx$
b) Sa se calculeze $\int^{e}_{1} f\left(x\right) dx$
c) Sa se determine ratia progresiei aritmetice avand termenul general $I_{n}=\int^{e^{n+1}}_{e^{n}}\left(f\left(x\right)-x\right) dx, n\geq 1$
Solutie:
Incepem prin a calcula integrala
a) $\int^{e}_{1}\left(\frac{\ln x}{x}+x-\frac{\ln x}{x}\right)dx=\int^{e}_{2} xdx=\frac{x^{2}}{2}|^{e}_{1}=\frac{e^{2}}{2}-\frac{1}{2}=\frac{e^{2}-1}{2}$ .
b) $\int^{e}_{1]}f\left(x\right) dx=\int^{e}_{1}\left(\frac{\ln x}{x}+x\right)dx=$
$\frac{\ln^{2} x}{2}|^{e}_{1}+\frac{x^{2}}{2}|^{e}_{1}=$
$\frac{\ln^{2} e}{2}-\frac{\ln^{2} 1}{2}+\frac{e^{2}}{2}-\frac{1}{2}=$
$\frac{1}{2}-0+\frac{e^{2}-1}{2}=\frac{1}{2}+\frac{e^{2}-1}{2}=$
$\frac{1+e^{2}-1}{2}=\frac{e^{2}}{2}$
Deci avem ca
$\int^{e}_{1}\left(\frac{\ln x}{x}\right)dx=\ln x\cdot\ln x|^{e}_{1}-\int^{e}_{1}\frac{\ln x}{x}dx\Rightarrow \int^{e}_{1}\frac{\ln x}{x}+\int^{e}_{1}\frac{\ln x}{x}=\ln^{2} x|^{e}_{1}\Rightarrow 2\int^{e}_{1}\frac{\ln x}{x}dx =\ln^{2} x|^{e}_{1}\Rightarrow \int^{e}_{1}\frac{\ln x}{x}dx=\frac{\ln^{2} x}{2}|^{e}_{1}$
Observam ca am luat separat si am calculat integrala $\int^{e}_{1}\frac{\ln x}{x} dx$ , am folosit metoda integrarii prin parti, adica am luat $f^{'}\left(x\right)=\ln x$ , iar $g\left(x\right)=\ln x$ , iar prin aplicarea si partea celei de-a doua a integrarii prin parti obtinem integrala de la care am plecat si astfel integrala care am gasit-o am trecut-o cu semn schimbat in partea stanga si astfel am obtinut de doua ori integrala de mai sus si astfel daca impartim prin 2 obtinem integrala de mai sus.
c) $I_{n}=\int^{e^{n+1}}_{e^{n}}\left(f\left(x\right)-x\right)dx=\int^{e^{n+1}}_{e^{n}}\left(\frac{\ln x}{x}+x-x\right)dx=\int^{e^{n+1}}_{e^{n}}\frac{\ln x}{x}dx$
Notam $\ln x=t \\ \frac{1}{x}dx=1\cdot dt \\ x=e^{n+1}\Rightarrow \ln e^{n+1}=t\Rightarrow n+1\cdot \ln e=t\Rightarrow n+1=t \\x=e^{n}\Rightarrow \ln e^{n}=t\Rightarrow n\cdot \ln e=t\Rightarrow t=n$
Deci integrala devine:
$\int^{e^{n+1}}_{e^{n}}\frac{\ln x}{x}dx=\int^{n+1}_{n}t dt=\frac{t^{2}}{2}|^{n+1}_{n}=\frac{\left(n+1\right)^{2}}{2}-\frac{n^{2}}{2}=\frac{\left(n+1\right)^{2}-n^{2}}{2}=\frac{n^{2}+2n+1-n^{2}}{2}=\frac{2n+1}{2}$
Am gasit ca $I_{n}=\frac{2n+1}{2}$
Acum daca calculam
$I_{1}=\frac{2\cdot 1+1}{2}=\frac{2+1}{2}=\frac{3}{2} \\I_{2}=\frac{2\cdot 2+1}{2}=\frac{5}{2}$
I_{2}-I_{1}=\frac{5}{2}-\frac{3}{2}=\frac{2}{2}=1$
Deci am gasit ca r=1.
Mai calculam
$I_{n+1}-I_{n}=\frac{2\left(n+1\right)+1}{2}-\frac{2n+1}{2}=\frac{2n+2+1}{2}-\frac{2n+1}{2}=\frac{2n+3-2n-1}{2}=\frac{2}{2}=1$ constant.
Si astfel gasim ca $I_{n}$ este progresie aritmetica cu ratia 1.