Write the first 4 terms of the sequence a n = n 2 ⋅ ( n − 1 ) ! a_{n}=n^2\cdot\left(n-1\right)!

Here we're asked to write the first four terms of the sequence a sub n is equal to n squared times n minus 1 factorial. Now it's perfectly fine for there to be a factorial in a sequence and we're just gonna treat this just as we would any other sequence by plugging in n equals 1 through n equals 4 in order to get those first four terms. So let's go ahead and get started with n equals 1. Now plugging n equals 1 into our formula here, I'm going to get a sub one is equal to 1 squared times 1 minus 1 factorial. Now simplifying this, 1 squared is just going to give me 1 and then 1 minus one is going to give me 0, so this is times 0 factorial. Now remember that 0 factorial is just equal to 1, so this ends up being 1 times 1 which is just 1. So my very first term, a sub 1, is just 1. Now for n equals 2, plugging that in, I get a sub 2 is equal to 2 squared times 2 minus 1 factorial. Now simplifying this, we get 2 squared, which is 4, and then that's multiplied by 2 minus 1 factorial, which will just give me 1 factorial. Now, 1 factorial is again just one, so I'm left here with 4 times 1, which is 4. So my second term in the sequence is 4. Now for our 3rd term, plugging in n equals 3, I get a sub 3, and that's equal to 3 squared times 3 minus 1 factorial. Now 3 squared is going to give us 9, and then 3 minus 1 factorial is 2 factorial, so this is 9 times 2 factorial. Now 2 factorial is just 2 times 1, so that's just 2. So this gives us 9 times 2, which is 18. So my 3rd term here is 18. Now for my 4th and final term that we want to find here, n equals 4. Plugging that in, I get a sub 4 is equal to 4 squared times 4 minus 1 factorial. So that's equal to 4 squared, which is 16. 4 minus 1 factorial gives me 3 factorial. So I have 16 times 3 factorial. Now 3 factorial is 3 times 2 times 1 which is equal to 6. So here I get 16 times 6 which will give me 96. So my 4th and final term here is 96. So my first four terms in this sequence are 1, 4, 18, and 96. Of course this will continue on but those are our first four terms. Thanks for watching, and I'll see you in the next one.

10. Combinatorics & Probability

Factorials

College Algebra

10. Combinatorics & Probability

Factorials

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Multiple Choice

Write the first 4 terms of the sequence $a_{n} = n^{2} \cdot (n - 1)!$

{1, 4, 18, 96}

{1, 2, 6, 24}

{0, 4, 18, 96}

{0, 2, 6, 24}

Verified step by step guidance

Understand the sequence formula: a_n = n^2 \cdot (n-1)!. This means for each term, you will square the term number n and multiply it by the factorial of (n-1).

Calculate the first term (n=1): a_1 = 1^2 \cdot (1-1)! = 1^2 \cdot 0! = 1 \cdot 1 = 1. Remember that 0! is defined as 1.

Calculate the second term (n=2): a_2 = 2^2 \cdot (2-1)! = 4 \cdot 1! = 4 \cdot 1 = 4.

Calculate the third term (n=3): a_3 = 3^2 \cdot (3-1)! = 9 \cdot 2! = 9 \cdot 2 = 18.

Calculate the fourth term (n=4): a_4 = 4^2 \cdot (4-1)! = 16 \cdot 3! = 16 \cdot 6 = 96.

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Struggling with College Algebra?

Write the first 4 terms of the sequence an=n2⋅(n−1)!a_{n}=n^2\cdot\left(n-1\right)!

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Factorials practice set

Write the first 4 terms of the sequence $a_{n} = n^{2} \cdot (n - 1)!$