Question

Write the first five terms of the sequence whose nth term is an = n (n+1)!.

Answer 1

2, 12, 72, 480 and 3600 (option A)

Step-by-step explanation:

`\begin{gathered} nth\text{ term ia given by:} \\ a_n\text{ = n(n + 1)!} \end{gathered}`

To get the first 5 terms, we will substitue numbers 1-5 for n

when n = 1

`\begin{gathered} a_1=1(1+1)!=1(2)! \\ a_1=\text{ 2! = 2}*1\text{ = 2} \end{gathered}`

when n = 2

`\begin{gathered} a_2=\text{ 2(2+1)! = 2(3)!} \\ a_2\text{ = 2}*3*2*1\text{ =12} \end{gathered}`

when x = 3

`\begin{gathered} a_3\text{ = 3(3+1)! = 3(4)!} \\ a_3\text{ = 3}*4*3*2*1 \\ a_3\text{ = }72 \end{gathered}`

when x = 4

`\begin{gathered} a_4\text{ = 4(4+1)! = 4(5)!} \\ a_4\text{ = 4}*5*4*3*2*1\text{ } \\ a_4\text{ = 4}80 \end{gathered}`

when x = 5

`\begin{gathered} a_5\text{ = }5(5+1)! \\ a_5\text{ = }5(6)!\text{ = 5}*6*5*4*3*2*1 \\ a_5\text{ = }3600 \end{gathered}`

Hence, the first 5 terms are:

2, 12, 72, 480 and 3600 (option A)