, p

N este :

A. + oo

B. 0

C. e

D.

E.

(eroare: eq.0/50641)$\lim_{x\to\0}\frac {\sum_{k=1}^m arctg \k^2x}{\sum_{k=1}^m \ln(1+k^3x)} $

A. (eroare: eq.1/50641)$ \frac {m(m+1)}{m+2} $

B.(eroare: eq.2/50641)$ \frac {2}{3}\cdot\frac{2m+1}{m(m+1)} $

C.(eroare: eq.3/50641)$ \frac {(m+1)(2m+1)}{2m^2} $

D. 0

E.(eroare: eq.4/50641)$ \frac {pi}{2e} $

=

, n

N\{0}

Incercari cu Nelu:

a patrat minus b patrat e asa:

[Citat] = , n N\{0}

Explicit:
[ equation]$x_n$[ /equation] = [ equation]$(1+\frac{1}{(n+1)^t})$[ /equation][ equation]$(1+\frac{1}{(n+2)^t})$[ /equation][ equation]$\dots$[ /equation][ equation]$(1+\frac{1}{(2n)^t})$[ /equation] , n[ equation]$\in$[ /equation] N\{0}

ASA CEVA NU SE FACE.
ATATA IESIT SI INTRAT DIN MATEMATICA STRICA PRIVIRII...

Ce este mult mai bine:
[ equation]
$$
x_n =
(1+\frac{1}{(n+1)^t})
(1+\frac{1}{(n+2)^t})
\dots
(1+\frac{1}{(2n)^t})
\ , \ n\in \N\setminus \{ 0 \}\ .
$$%
[ /equation]

Iata cum se compileaza cele de mai sus:

Si mai bine, deoarece ma supara parantezele:
[ equation]
$$
x_n =
\left( 1+\frac{1}{(n+1)^t} \right)
\left( 1+\frac{1}{(n+2)^t} \right)
\dots
\left( 1+\frac{1}{(2n)^t} \right)
\ , \
n\in \N\setminus \{ 0 \}\ .
$$%
[ /equation]

Iata cum se compileaza cele de mai sus:

[equation]Deci datele de pornire sunt:

$a_0=2;a_1=16; a_{n+1}^2=a_n\cdot a_{n-1},\forall n\geq 1.$

Pentru $n=1$ avem $a_2^2=a_1\cdot a_0$ .

Pentru $n=2$ avem $a_3^2=a_2\cdot a_1$ .

Pentru $n=3$ avem $a_4^2=a_3\cdot a_2$

$\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots$

Atribuind lui $n$ valoarea $n-2$ avem $a_{n-1}^2=a_{n-2}\cdot a_{n-3}$

Atribuind lui $n$ valoarea $n-1$ avem $a_n^2=a_{n-1}\cdot a_{n-2}$

Atribuind lui $n$ valoarea $n$ avem $a_{n+1}^2=a_n\cdot a_{n-1}$

A se da un clip pe acel [Citeaza]...