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pogonyaev
3 years ago
12

Help with math problem- finding first four terms of a sequence with a factorial. An=2(n+1)!

Mathematics
1 answer:
earnstyle [38]3 years ago
7 0

a_1=2\cdot(1+1)!=2\cdot2!=2\cdot1\cdot2=\boxed{4}\\\\ a_2=2\cdot(2+1)!=2\cdot3!=2\cdot1\cdot2\cdot3=\boxed{12}\\\\ a_3=2\cdot(3+1)!=2\cdot4!=2\cdot1\cdot2\cdot3\cdot4=\boxed{48}\\\\ a_4=2\cdot(4+1)!=2\cdot5!=2\cdot1\cdot2\cdot3\cdot4\cdot5=\boxed{240}

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Determine if the sequence is arithmetic. If it is, find the common difference. Is the sequence a function ?
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6)\\r=a_{n+1}-a_n\to r=a_2-a_1=a_3-a_2=a_4-a_3=...\\\\a_1=-7;\ a_2=-10;\ a_3=-11;\ a_4=-13\\\\a_2-a_1=-10-(-7)=-10+7=-3\\a_3-a_2=-11-(-10)=-11+10=-1\\\\a_3-a_2\neq a_2-a_1\\\\\text{It's not an arithmetic sequence}


\r=\dfrac{a_{n+1}}{a_n}\\\\r=\dfrac{a_2}{a_1}=\dfrac{a_3}{a_2}=\dfrac{a_4}{a_3}=...\\\\8.)\\a_1=4;\ a_2=20;\ a_3=100;\ a_4=500\\\\r=\dfrac{20}{4}=\dfrac{100}{20}=\dfrac{500}{100}=5\\\\10.)\\a_1-30;\ a_2=-45;\ a_3=-50;\ a_4=-65\\\\\dfrac{a_2}{a_1}=\dfrac{-45}{-30}=\dfrac{3}{2}\\\\\dfrac{a_3}{a_2}=\dfrac{-50}{-45}=\dfrac{10}{9}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}
12.)\\a_1=-7;\ a_2=-14;\ a_3=-21;\ a_4=-28\\\\\dfrac{a_2}{a_1}=\dfrac{-14}{-7}=2\\\\\dfrac{a_3}{a_2}=\dfrac{-21}{-14}=\dfrac{3}{2}\\\\\dfrac{a_2}{a_1}\neq\dfrac{a_3}{a_2}\\\\\text{It's not a geometric sequence}\\\\\text{It's a function}


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