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)

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

$\begin{gathered} a_1=1(1+1)!=1(2)! \\ a_1=\text{ 2! = 2}\times1\text{ = 2} \end{gathered}$

when n = 2

$\begin{gathered} a_2=\text{ 2(2+1)! = 2(3)!} \\ a_2\text{ = 2}\times3\times2\times1\text{ =12} \end{gathered}$

when x = 3

$\begin{gathered} a_3\text{ = 3(3+1)! = 3(4)!} \\ a_3\text{ = 3}\times4\times3\times2\times1 \\ a_3\text{ = }72 \end{gathered}$

when x = 4

$\begin{gathered} a_4\text{ = 4(4+1)! = 4(5)!} \\ a_4\text{ = 4}\times5\times4\times3\times2\times1\text{ } \\ a_4\text{ = 4}80 \end{gathered}$

when x = 5

$\begin{gathered} a_5\text{ = }5(5+1)! \\ a_5\text{ = }5(6)!\text{ = 5}\times6\times5\times4\times3\times2\times1 \\ a_5\text{ = }3600 \end{gathered}$

Hence, the first 5 terms are:

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