# Find the first four terms of a sequence that has the nth term. {eq}a_{n} = 2(n + 1)! {/eq}

## Question:

Find the first four terms of a sequence that has the nth term.

{eq}a_{n} = 2(n + 1)! {/eq}

## Sequence:

We know that in a set if some numbers are present then all of the numbers be unique. When this set of numbers follow a particular pattern then set of numbers are in sequence. This sequence may be finite or infinite. Now, the {eq}n^{th} {/eq} term of this sequence is called general term and from this term, we can find any terms of that sequence.

The {eq}n^{th} {/eq} term of a sequence is given by:

{eq}a_n = 2(n+1)! {/eq}

Now, substituting {eq}n = 1, \, 2, \, 3 \, \text{and} \, 4 {/eq} into the given term, we have:

{eq}\begin{align*} a_1 &= 2(1+1)! \\[0.3 cm] &= 2(2)! \\[0.3 cm] &= 2(2 \times 1 ) & \left[ \because n! = n(n-1)(n-2) ..... 1 \right] \\[0.3 cm] &= 4 \end{align*} {/eq}

{eq}\begin{align*} a_2 &= 2(2+1)! \\[0.3 cm] &= 2(3)! \\[0.3 cm] &= 2(3 \times 2 \times 1 ) & \left[ \because n! = n(n-1)(n-2) ..... 1 \right] \\[0.3 cm] &= 12 \end{align*} {/eq}

{eq}\begin{align*} a_3 &= 2(3+1)! \\[0.3 cm] &= 2(4)! \\[0.3 cm] &= 2(4 \times 3 \times 2 \times 1 ) & \left[ \because n! = n(n-1)(n-2) ..... 1 \right] \\[0.3 cm] &= 48 \end{align*} {/eq}

{eq}\begin{align*} a_4 &= 2(4+1)! \\[0.3 cm] &= 2(5)! \\[0.3 cm] &= 2(5 \times 4 \times 3 \times 2 \times 1 ) & \left[ \because n! = n(n-1)(n-2) ..... 1 \right] \\[0.3 cm] &= 240 \end{align*} {/eq}

Hence, the first four terms of the sequence are {eq}\color{blue}{\boxed{ 4, \, 12, \, 48 \, \text{and} \, 240 }} {/eq}