Question

Write down the first five terms of the sequence an=(n+4)!2n2+6n+7a1 = a2 = a3 = a4 = a5 =

Answer 1

Step 1

Given;

$\begin{gathered} a_n=\frac{(n+4)!}{2n^2+6n+7} \\ n=1,2,3,4,5 \end{gathered}$

Step 2

$a_1=\frac{(1+4)!}{2(1)^2+6(1)+7}=\frac{120}{15}=8$$a_2=\frac{(2+4)!}{2(2)^2+6(2)+7}=\frac{720}{27}=\frac{80}{3}$$a_3=\frac{(3+4)!}{2(3)^2+6(3)+7}=\frac{5040}{43}$$a_4=\frac{(4+4)!}{2(4)^2+6(4)+7}=\frac{8!}{63}=640$$a_5=\frac{(5+4)!}{2(5)^2+6(5)+7}=\frac{9!}{87}=\frac{120960}{29}$