a_1&=-2
a_(n+1)&=5a_n+2n
To do so, we will use a table. To find a_2, we will substitute 1 for n in the above formula. To find a_3, we will substitute 2 for n, and so on.
n
a_(n+1)=5a_n+n
5a_n+n
a_(n+1)
-
a_1= -2
-
-
1
a_(1+1)=5a_1+2( 1) ⇕ a_2=5a_1+2( 1)
a_2=5 a_1+2 ⇓ a_2=5( -2)+2
a_2= -8
2
a_(2+1)=5a_2+2( 2) ⇕ a_3=5a_2+2( 2)
a_3=5 a_2+4 ⇓ a_3=5( -8)+4
a_3= -36
3
a_(3+1)=5a_3+2( 3) ⇕ a_4=5a_3+2( 3)
a_4=5 a_3+6 ⇓ a_4=5( -36)+6
a_4= -174
4
a_(4+1)=5a_4+2( 4) ⇕ a_5=5a_4+2( 4)
a_5=5 a_4+8 ⇓ a_5=5( -174)+8
a_5= -862
The first five terms of the sequence are -2, -8, -36, -174, and -862.
