Question

se da sirul:12,20,30,42,56,......Află nr.de pe locul 2009.

Answer 1

12,20,30,42,56...

$a_1 = \bold{12} \\ a_2 = a_1+8 = a_1+ 2 \cdot 2 = \bold{a_1+2\cdot( 2+2)}\\ a_3 = a_2+10= a_2+2\cdot 5 = \bold{a_2+2\cdot (3+2)}\\ a_4 = a_3+12 = a_3+ 2\cdot 6 = \bold{a_3+2\cdot (4+2)} \\ a_5 = a_4 +14 = a_4 +2\cdot 7 = \bold{a_4+2\cdot (5+2)}\\ ... \\ a_n = \bold{a_{n-1} + 2\cdot (n+2)}\\ \\ ---------------------(+) \\ \\ \Rightarrow a_1 +a_2+...+a_n = 12 + a_1+ a_2 +...+ a_{n-1}+ \\ +2\cdot(2+2)+ 2\cdot (3+2)+...+2\cdot (n+2)$

$\Rightarrow a_1+a_2+...+a_n-(a_1+a_2+...+a_{n-1}) = \\ = 12+2\cdot \Big[(2+2)+(3+2)+...+(n+2)\Big] \\ \\ \Rightarrow a_n = 12+2\cdot \Big[(2+3+...+n)+\underset{de~n-1~ori}{\underbrace{(2+2+...+2)}}\Big] \\ \\ \Rightarrow a_n = 12+2\cdot \Big[(1+2+3+...+n) - 1 + 2\cdot (n-1)\Big] \\ \\ \Rightarrow a_n = 12+ 2\cdot \Big[\dfrac{n(n+1)}{2} -1 + 2(n-1)\Big] \\ \Rightarrow a_n = 12 +\Big[n(n+1)-2+4(n-1)\Big] \\ \\ \Rightarrow a_n = 10+ n(n+1) + 4(n-1)$

$\\ \\ \Rightarrow a_n = 10 + n(n+1)+4n-4 = 6 +n\big[(n+1)+4\big] \\ \\ \Rightarrow \boxed{a_n = 6 + n(n+5)} \\ \\ \Rightarrow a_{2009} = 6 + 2009\cdot (2009+5) = 6 + 2009 \cdot 2014 = 4046132$

=> Numărul de pe locul 2009 este 4046132.

Answer 2

Am atasat o rezolvare.