Question

Demonstrati ca sirul definit prin formula:
$c_{n} = \frac{ 3_{n}+4}{n+2}$ este strict crescator. UGEENT!

Answer 1

 c_n  ≤ c{n+1                cu n∈ N
(3n +4 ) ( n+2 ) ≤ [ 3(  n+1 )  +4 ]  /   [ ( n+1 )  +2 ]
( 3n +4) /( n +2) ≤ ( 3n +7 ) / ( n+3 ) 
( 3n +4) /( n +2)   -    ( 3n +7 ) / ( n+3 )  ≤ 0 
( 3n² +4n+9n +12 - 3n² -7n-6n -14 )  /  [ ( n+2)·( n+3) ] ≤ 0
( 13 n +12 - 13n -14 )  /  [ ( n+2)·( n+3) ]  ≤ 0
(  -2 )  / [ ( n+2)· ( n+3 ) ] ≤ 0 adevarat  cu n∈ N numitorul este pozitiv 
dar numaratorul= -2           ⇒     c_n  ≤ c{n+1}     monoton crescator