Answer :
[tex]a_n=n(n-2)-3 \\ \\
a_1=1(1-2)-3=1 \times (-1)-3=-1-3=-4 \\
a_2=2(2-2)-3=2 \times 0-3=0-3=-3 \\
a_3=3(3-2)-3= 3 \times 1-3=3-3=0 \\
a_4=4(4-2)-3=4 \times 2 -3 =8-3 =5
[/tex]
The first four terms are -4, -3, 0, 5.
The first four terms are -4, -3, 0, 5.
Answer:
As per the statement:
Given a sequence:
[tex]a_n = n(n-2)-3[/tex]
where, n is the number of terms;
We have to find the first four terms of this sequence:
For n =1
[tex]a_1 = 1(1-2)-3 =1(-1)-3 = -1-3 = -4[/tex]
⇒[tex]a_1 = -4[/tex]
For n =2
[tex]a_2= 2(2-2)-3 =2(0)-3 = 0-3 = -3[/tex]
⇒[tex]a_2 = -3[/tex]
For n =3
[tex]a_3 = 3(3-2)-3 =3(1)-3 = 3-3 = 0[/tex]
⇒[tex]a_3= 0[/tex]
For n =4
[tex]a_4 = 4(4-2)-3 =4(2)-3 = 8-3 = 5[/tex]
⇒[tex]a_4= 5[/tex]
Therefore, the first four terms of the sequence are:
-4, -3, 0 , 5
