a_1&=12
a_(n+1)&=a_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)=a_n+n
a_n+n
a_(n+1)
-
a_1= 12
-
-
1
a_(1+1)=a_1+ 1 ⇕ a_2=a_1+ 1
a_2= a_1+1 ⇓ a_2= 12+1
a_2= 13
2
a_(2+1)=a_2+ 2 ⇕ a_3=a_2+ 2
a_3= a_2+2 ⇓ a_3= 13+2
a_3= 15
3
a_(3+1)=a_3+ 3 ⇕ a_4=a_3+ 3
a_4= a_3+3 ⇓ a_4= 15+3
a_4= 18
4
a_(4+1)=a_4+ 4 ⇕ a_5=a_4+ 4
a_5= a_4+4 ⇓ a_5= 18+4
a_5= 22
The first five terms of the sequence are 12, 13, 15, 18, and 22.
