a)
[tex]\it Se \ poate \ observa\ c\breve a\ termenul \ general\ al\ \d sirului\ este:\\ \\ a_n=n^2(n+1)\\ \\ a_5=5^2(5+1)=25\cdot6=150\\ \\ a_6=6^2(6+1)=36\cdot7=252\\ \\ a_7=7^2(7+1)=49\cdot8=392[/tex]
b)
[tex]\it a_n=n^2(n+1)=n\cdot n(n+1)=\ par, \ deoarece\ n(n+1)=par,\ \forall\ n\in\mathbb N[/tex]
c)
[tex]\it 1452=2^2\cdot3\cdot11^2=11^2\cdot12=11^2(11+1) \Rightarrow da, este \ al\ 11-lea\ termen\\ \\ \\ 3700=100\cdot37=10^2\cdot37\ne10^2(10+1),\ deci\ nu\ apar\c{\it t}ine\ \d sirului\ .[/tex]
