Verified Answer :

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\(a_n=n(n-2)-3 \\ \\ a_1=1(1-2)-3=1 \times (-1)-3=-1-3=-4 \\ a_2=2(2-2)-3=2 \times 0-3=0-3=-3 \\ a_3=3(3-2)-3= 3 \times 1-3=3-3=0 \\ a_4=4(4-2)-3=4 \times 2 -3 =8-3 =5\)

The first four terms are -4, -3, 0, 5.

Answer:

As per the statement:

Given a sequence:

\(a_n = n(n-2)-3\)

where, n is the number of terms;

We have to find the first four terms of this sequence:

For n =1

\(a_1 = 1(1-2)-3 =1(-1)-3 = -1-3 = -4\)

⇒\(a_1 = -4\)

For n =2

\(a_2= 2(2-2)-3 =2(0)-3 = 0-3 = -3\)

⇒\(a_2 = -3\)

For n =3

\(a_3 = 3(3-2)-3 =3(1)-3 = 3-3 = 0\)

⇒\(a_3= 0\)

For n =4

\(a_4 = 4(4-2)-3 =4(2)-3 = 8-3 = 5\)

⇒\(a_4= 5\)

Therefore, the first four terms of the sequence are:

-4, -3, 0 , 5

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