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Chapter 8: Problem 21

The general term of a sequence is given and involves a factorial. Write the first four terms of each sequence. $$a_{n}=2(n+1) !$$

Short Answer

Expert verified
The first four terms of the sequence are 4, 12, 48, and 240.

Step by step solution

01

Calculate the First Term

Substitute n=1 into the general term \(a_{n}=2(n+1) !\). This will give \(a_1=2(1+1)!\), which simplifies to 4 (because 2! = 2 and multiplying by 2 gives 4). So, the first term, \(a_1\) is 4.
02

Calculate the Second Term

Substitute n=2 into the general term \(a_{n}=2(n+1) !\). This will give \(a_2=2(2+1)!\), which simplifies to 12 (because 3! = 6 and multiplying by 2 gives 12). Therefore, the second term, \(a_2\) is 12.
03

Calculate the Third Term

Substitute n=3 into the general term \(a_{n}=2(n+1) !\). Hence, \(a_3=2(3+1)!\), this simplifies to 48 (because 4! = 24, and multiplying by 2 gives 48). Therefore, the third term \(a_3\) is 48.
04

Calculate the Fourth Term

Substitute n=4 into the general term \(a_{n}=2(n+1) !\). This gives \(a_4=2(4+1)!\), which simplifies to 240 (because 5! = 120 and multiplying by 2 gives 240). Therefore, the fourth term \(a_4\) is 240.

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