Question

write down the first five terms of the sequence $a_n = \frac{(n + 1)!}{2n^2 + 4n + 6}$
$a_1 = $
$a_2 = $
$a_3 = $
$a_4 = $
$a_5 = $
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Explanation:

Step1: Find \(a_1\)

Substitute \(n = 1\) into \(a_n=\frac{(n + 1)!}{2n^2+4n + 6}\).
\((1 + 1)! = 2! = 2\), \(2(1)^2+4(1)+6 = 2 + 4 + 6 = 12\).
So \(a_1=\frac{2}{12}=\frac{1}{6}\).

Step2: Find \(a_2\)

Substitute \(n = 2\) into the formula.
\((2 + 1)! = 3! = 6\), \(2(2)^2+4(2)+6 = 8 + 8 + 6 = 22\)? Wait, no: \(2(4)+8 + 6 = 8 + 8 + 6 = 22\)? Wait, no, \(2n^2+4n + 6\) when \(n=2\): \(2(4)+4(2)+6=8 + 8 + 6 = 22\)? Wait, no, wait \(2n^2+4n + 6 = 2(n^2 + 2n + 3)\). For \(n=2\), \(n^2+2n + 3=4 + 4 + 3=11\), so \(2\times11 = 22\)? Wait, but \((2 + 1)! = 6\), so \(a_2=\frac{6}{22}=\frac{3}{11}\)? Wait, no, wait I miscalculated \(2n^2+4n + 6\) for \(n=2\): \(2(2)^2=8\), \(4(2)=8\), \(8 + 8 + 6 = 22\). Yes. Then \((2 + 1)! = 6\), so \(a_2=\frac{6}{22}=\frac{3}{11}\). Wait, no, wait \((n + 1)!\) when \(n=2\) is \(3! = 6\), correct. Then denominator: \(2(2)^2+4(2)+6=8 + 8 + 6 = 22\). So \(a_2=\frac{6}{22}=\frac{3}{11}\).

Wait, no, wait let's recalculate denominator: \(2n^2+4n + 6 = 2(n^2 + 2n + 3)\). For \(n=1\): \(1 + 2 + 3 = 6\), \(2\times6=12\), correct. For \(n=2\): \(4 + 4 + 3 = 11\), \(2\times11=22\), correct. For \(n=3\): \(9 + 6 + 3 = 18\), \(2\times18=36\). \((3 + 1)! = 4! = 24\), so \(a_3=\frac{24}{36}=\frac{2}{3}\). Wait, that's better. Let's do \(n=3\):

Step3: Find \(a_3\)

\(n=3\): \((3 + 1)! = 4! = 24\), denominator: \(2(3)^2+4(3)+6 = 18 + 12 + 6 = 36\). So \(a_3=\frac{24}{36}=\frac{2}{3}\).

Step4: Find \(a_4\)

\(n=4\): \((4 + 1)! = 5! = 120\), denominator: \(2(4)^2+4(4)+6 = 32 + 16 + 6 = 54\). So \(a_4=\frac{120}{54}=\frac{20}{9}\).

Step5: Find \(a_5\)

\(n=5\): \((5 + 1)! = 6! = 720\), denominator: \(2(5)^2+4(5)+6 = 50 + 20 + 6 = 76\)? Wait, no: \(2(25)=50\), \(4(5)=20\), \(50 + 20 + 6 = 76\). Then \((5 + 1)! = 720\), so \(a_5=\frac{720}{76}=\frac{180}{19}\). Wait, no, wait \(2n^2+4n + 6\) for \(n=5\): \(2(25)+20 + 6 = 50 + 20 + 6 = 76\). \((5 + 1)! = 720\). So \(a_5=\frac{720}{76}=\frac{180}{19}\). Wait, but let's check \(n=3\) again: \(n=3\), denominator \(2(9)+12 + 6 = 18 + 12 + 6 = 36\), numerator \(4! = 24\), so \(24/36 = 2/3\), correct. \(n=4\): denominator \(2(16)+16 + 6 = 32 + 16 + 6 = 54\), numerator \(5! = 120\), \(120/54 = 20/9\), correct. \(n=5\): denominator \(2(25)+20 + 6 = 50 + 20 + 6 = 76\), numerator \(6! = 720\), \(720/76 = 180/19\), correct. \(n=1\): \(2! = 2\), denominator \(2(1)+4(1)+6 = 2 + 4 + 6 = 12\), \(2/12 = 1/6\), correct. \(n=2\): \(3! = 6\), denominator \(2(4)+8 + 6 = 8 + 8 + 6 = 22\)? Wait, no, wait \(2n^2+4n + 6\) when \(n=2\): \(2(4) = 8\), \(4(2)=8\), \(8 + 8 + 6 = 22\). So \(6/22 = 3/11\), correct.

Wait, but let's re-express the denominator: \(2n^2 + 4n + 6 = 2(n^2 + 2n + 3)\). Maybe factor \(n^2 + 2n + 3\), but it doesn't factor nicely. So proceeding:

\(a_1 = \frac{(1+1)!}{2(1)^2 + 4(1) + 6} = \frac{2!}{2 + 4 + 6} = \frac{2}{12} = \frac{1}{6}\)

\(a_2 = \frac{(2+1)!}{2(2)^2 + 4(2) + 6} = \frac{3!}{8 + 8 + 6} = \frac{6}{22} = \frac{3}{11}\)

\(a_3 = \frac{(3+1)!}{2(3)^2 + 4(3) + 6} = \frac{4!}{18 + 12 + 6} = \frac{24}{36} = \frac{2}{3}\)

\(a_4 = \frac{(4+1)!}{2(4)^2 + 4(4) + 6} = \frac{5!}{32 + 16 + 6} = \frac{120}{54} = \frac{20}{9}\)

\(a_5 = \frac{(5+1)!}{2(5)^2 + 4(5) + 6} = \frac{6!}{50 + 20 + 6} = \frac{720}{76} = \frac{180}{19}\)

Answer:

\(a_1 = \boldsymbol{\frac{1}{6}}\)
\(a_2 = \boldsymbol{\frac{3}{11}}\)
\(a_3 = \boldsymbol{\frac{2}{3}}\)
\(a_4 = \boldsymbol{\frac{20}{9}}\)
\(a_5 = \boldsymbol{\frac{180}{19}}\)