Răspuns:
4. n∈N , n≥2
3Cₙ¹ + 2Cₙ² =8
[tex]3* \frac{n!}{(n-1)!*1!} + 2*\frac{n!}{(n-2)!*2!} = 8\\\\<=> 3*\frac{n*(n-1)*(n-2)!}{(n-1)*(n-2)!} + 2* \frac{n*(n-1)*(n-2)!}{(n-2)!*2}=8\\\\<=> 3n + n*(n-1)=8 \\\\<=> 3n+n^{2}-n =8\\\\<=> n^{2} + 2n-8=0\\\\[/tex]
Δ= 4+4*8=36 >0
n₁= -2+6/2 = 2 ∈ [2;∞) ∩ N
n₂= -2-6/2 = -4 ∉ [2;∞) ∩ N
5. cf teoremei lui Ceva, dreptele AA', BB', CC' sunt concurente
<=> AC' / C'B * BA'/A'C * CB'/B'A = 1 --- (1)
A'C = 2* BA' ( rel. vectoriala) => A'C/ BA' =2 => BA'/ A'C = 1/2
B'C = 2/5 * AC ( rel.vectoriala) => CB'/ AC= 2/5 <=> CB'/ AC-CB' = 2/ 5-2
<=> CB'/ B'A = 2/3
C'A= 3* BC' ( rel.vectoriala) => AC'/ C'B= 3
revenind in (1)
avem 3* 1/2 * 2/3=1
<=> 1=1 ad ===> rel. adevarata => AA', BB', CC' concurente
