Question

Dau coroanaaa!!!Studiati monotonia sirului an=2n+1/n+4.

Answer 1

ca la clasele 7-:-9

(2n+8-7)/(n+4)=2-7/(n+4)
2 constant
n+4 , crescator
1/(n+4 descrescator
7/(n+4) descrescator
-7(n+4) crescator
constant + crescator=crescator

ca la cl; 11
f(x) extensie la R
f'(x)=(2(x+4)-2x-1)/(x+4)²=7/(x+4)² >0, ∀x∈R, f(x) crescatoare f(n) creste

ca la clasa a5-a
a(0)=1/4
a(1)=3/5 creste
a(2)=5/6
a(10)=21/14
a(100)=201/104
a(1000)=2001/1004
se ...."apropie" de 2

ca la cl a 9-a, cam asa   cum se cere
a(n+1)-a(n)=(2n+3)/(n+5)-(2n+1)/(n+4)=((2n+3)(n+4)-(2n+1)(n+5))/(n+5)(n+4)=
(2n²+11n+12-2n²-11n-5)/(n²+9n+20)=
7/(n²+9n+20)>0, ∀n∈N deci sirul este strict crescator
si la  metoda asta , ca si metoda 1, dar aici se vede mai bine ,observam :cresterea este din ce in ce mai mica

Answer 2

[tex]\text{Sau fara sa ne complicam:}\\ \dfrac{a_{n+1}}{a_n}=\dfrac{\frac{2n+3}{n+5}}{\frac{2n+1}{n+4}}=\dfrac{(2n+3)(n+4)}{(2n+1)(n+5)}=\dfrac{2n^2+3n+8n+12}{2n^2+10n+n+5}=\\ =\dfrac{2n^2+11n+12}{2n^2+11n+5} \ \textgreater \ \dfrac{2n^2+11n+5}{2n^2+11n+5}=1\Rightarrow a_{n+1}\ \textgreater \ a_{n}\\ \text{Deci sirul este crescator.} [/tex]