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

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

14. Sequences & Series

Review of Factorials

Business Calculus

14. Sequences & Series

Review of Factorials

Struggling with Business Calculus?

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Multiple Choice

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

{1, 4, 18, 96}

{1, 2, 6, 24}

{0, 4, 18, 96}

{0, 2, 6, 24}

Verified step by step guidance

Step 1: Understand the given sequence formula. The sequence is defined as a_n = n^2 * (n-1)!, where n is the term number, n^2 represents n squared, and (n-1)! is the factorial of (n-1).

Step 2: Calculate the first term (n=1). Substitute n=1 into the formula: a_1 = 1^2 * (1-1)!. Simplify (1-1)! to 0! (which equals 1 by definition of factorials). Then calculate 1^2 * 1.

Step 3: Calculate the second term (n=2). Substitute n=2 into the formula: a_2 = 2^2 * (2-1)!. Simplify (2-1)! to 1! (which equals 1). Then calculate 2^2 * 1.

Step 4: Calculate the third term (n=3). Substitute n=3 into the formula: a_3 = 3^2 * (3-1)!. Simplify (3-1)! to 2! (which equals 2). Then calculate 3^2 * 2.

Step 5: Calculate the fourth term (n=4). Substitute n=4 into the formula: a_4 = 4^2 * (4-1)!. Simplify (4-1)! to 3! (which equals 6). Then calculate 4^2 * 6.