[tex]\it a_1 =\ 0
\\
a_2 =\ 1
\\
a_3 =\ 4
\\
a_4 =10
\\
a_5 =20
\\
a_6 =35
\\
a_7 =56 \\
----
[/tex]
[tex]\it a_2-a_1= 1-0=1=\dfrac{2\cdot1}{2} = \dfrac{2\cdot(2-1)}{2} = \dfrac{2^2-2}{2}
\\ \\
a_3-a_2= 4-1=3=\dfrac{3\cdot2}{2} = \dfrac{3\cdot(3-1)}{2} = \dfrac{3^2-3}{2}
\\ \\
a_4-a_3= 10-4=6=\dfrac{4\cdot3}{2} = \dfrac{4\cdot(4-1)}{2} = \dfrac{4^2-4}{2}
\\ \\
a_5-a_4= 20-10=10=\dfrac{5\cdot4}{2} = \dfrac{5\cdot(5-1)}{2} = \dfrac{5^2-5}{2}
\\ .\\.\\ . \\
a_n-a_{n-1} = \dfrac{n^2-n}{2}
\\ ------------- \ adun\breve{a}m :
\\ \\
(a_2+a_3+a_4+a_5+...+a_n)-(a_1+a_2+a_3+a+4+...+a_{n-1})=[/tex]
[tex]\it \dfrac{1}{2}[(2^2+3^2+4^2+...+n^2)- (2+3+4+...+n)] \Rightarrow a_n-a_1= \\ \\ \\
= \dfrac{1}{2}[(1^2+2^2+3^2+4^2+...+n^2)- (1+2+3+4+...+n)] \Rightarrow a_n= \\ \\ \\
=
\dfrac{1}{2}\left[\dfrac{n(n+1)(2n+1)}{6} -\dfrac{n(n+1)}{2}\right] =\dfrac{1}{4}n(n+1)\left(\dfrac{2n+1}{3}-1\right)=[/tex]
[tex]\it= \dfrac{1}{4}n(n+1)\cdot\dfrac{2n+1-3}{3} = \dfrac{1}{4}n(n+1)\cdot\dfrac{2n-2}{3} =
\\ \\ \\
=\dfrac{1}{4}n(n+1)\cdot\dfrac{2(n-1)}{3} = \dfrac{(n-1)n(n+1)}{6} [/tex]
[tex]\it Deci,\ a_n = \dfrac{(n-1)n(n+1)}{6} [/tex]
[tex]\it a_8 = \dfrac{7\cdot8\cdot9}{6} = 84
\\ \\ \\
a_9 = \dfrac{8\cdot9\cdot10}{6} = 120
\\ \\ \\
a_{15} = \dfrac{14\cdot15\cdot16}{6} = 560[/tex]
[tex]\it a_n=\dfrac{(n-1)n(n+1)}{6} \Rightarrow 6\cdot a_n= (n-1)n(n+1)
\\ \\ \\
6\cdot2017 = 2\cdot3\cdot2017 \neq (n-1)n(n+1)
[/tex]
Așadar, 2017 nu este un termen al șirului.
