Question

write the first four terms of the sequence whose general term is given below.
$a_{n} = 2(n + 1)!$
$a_{1} = \square$

Explanation:

Step1: Identify n for \(a_1\)

For the first term \(a_1\), \(n = 1\).

Step2: Substitute n into the formula

Substitute \(n = 1\) into \(a_n=2(n + 1)!\). We know that \(n=1\), so \(n + 1=2\), and \(2! = 2\times1=2\). Then \(a_1=2\times(1 + 1)!=2\times2!=2\times2 = 4\).

Step3: Find \(a_2\)

For \(a_2\), \(n = 2\). Then \(n + 1=3\), \(3! = 3\times2\times1 = 6\). So \(a_2=2\times(2 + 1)!=2\times3!=2\times6 = 12\).

Step4: Find \(a_3\)

For \(a_3\), \(n = 3\). Then \(n + 1=4\), \(4! = 4\times3\times2\times1=24\). So \(a_3=2\times(3 + 1)!=2\times4!=2\times24 = 48\).

Step5: Find \(a_4\)

For \(a_4\), \(n = 4\). Then \(n + 1=5\), \(5! = 5\times4\times3\times2\times1 = 120\). So \(a_4=2\times(4 + 1)!=2\times5!=2\times120 = 240\).

Answer:

The first four terms are \(a_1 = 4\), \(a_2=12\), \(a_3 = 48\), \(a_4=240\). (If only \(a_1\) is required, the answer is \(4\))