a_1&=-4
a_(n+1)&=2a_n+n
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)=2a_n+n
2a_n+n
a_(n+1)
-
a_1= -4
-
-
1
a_(1+1)=2a_1+ 1 ⇕ a_2=2a_1+ 1
a_2=2 a_1+1 ⇓ a_2=2( -4)+1
a_2= -7
2
a_(2+1)=2a_2+ 2 ⇕ a_3=2a_2+ 2
a_3=2 a_2+2 ⇓ a_3=2( -7)+2
a_3= -12
3
a_(3+1)=2a_3+ 3 ⇕ a_4=2a_3+ 3
a_4=2 a_3+3 ⇓ a_4=2( -12)+3
a_4= -21
4
a_(4+1)=2a_4+ 4 ⇕ a_5=2a_4+ 4
a_5=2 a_4+4 ⇓ a_5=2( -21)+4
a_5= -38
The first five terms of the sequence are -4, -7, -12, -21, and -38.
