Compute the first four partial sums and find a formula for the nthn^{\operatorname{th}}nth partial sum.∑n=1∞2n−1\sum_{n=1}^{\infty}2n-1∑n=1∞2n−1

1,4,9,16;Sn=n21,4,9,16;S_{n}=$$n^21$$,4,9,16;Sn=n2

Compute the first four partial sums and find a formula for the n th n^{\operatorname{th}} n th partial sum. ∑ n = 1 ∞ 2 n − 1 \sum_{n=1}^{\infty}2n-1 ∑ n = 1 ∞ 2 n − 1

This problem, we are asked to compute the first four partial sums and find a formula for the nth partial sum based on a sequence that we're given. Now what I can see right here is this would be our sequence a sub n. Now I can use this to figure out what the first four partial sums are and the nth partial sum. And I think the most straightforward way to do this is if I create kind of a table here. So what I'm gonna do is say that my a sub n is clearly what we have in this problem, the two n minus one. But What I'm going to do is write out the first four pieces of our sequence right here. So we're going to have a sub one. Well, a sub one is just gonna be two times one minus one, which is just equal to one. Now let's move to a sub two. Ace of two will be two times two minus one, which is the same thing as four minus one, or three. Now let's move to ace of three. Ace of three is going to be two times three minus one, which is the same thing as six minus one, or five. Now let's move to ace of four. Ace of four is the same thing as two times four minus one, which is eight minus one or seven. So now I've wrote out the first four parts of my sequence right here. But I'm not looking for the sequence. I'm looking for the finite series, also known as the partial sum. So what I'm looking for here is s of n, because that's ultimately where I'm going to end up with this problem. We want to find a formula for the nth partial sum. But before I go ahead and do that, I think it's going to be best if I write out a few terms to see if there's a pattern I can recognize for our partial sum. So let's start with s of one. Well, s of one, we know, correlates directly to a sub one, so that's just gonna be one. Now let's move to s of two. Well, s of two is just gonna be a sub one plus a sub two, which is the same thing as one plus three, or four. Now let's move to s of three. Well, s of three is the same thing as a one plus a two plus a sub three. So we have one plus three plus five, which all comes out to nine. And now let's last move to ace or s of four, I should say. S of four is going to be adding these first four terms in our sequence, which is going to be one plus three plus five plus seven, which all comes out to 16. So this is what happens when we go to s of one, s of two, s of three, s of four. And that is the first four partial sums. And if I want to find a formula for the nth partial sum, what I need to do is see if I can recognize a pattern for each of these values that we got here that correlates to the index of our partial sum. Well, looking at this, notice that we got one at the start. This is the same thing as just writing this as one squared. Now there's a reason that I'm doing this here, because notice that four is the same thing as two squared. Nine is the same thing as three squared. 16 is the same thing as four squared. Notice that each of these numbers that are being squared corresponds to each index that we have in our finite series. So if I want to write s of n here, that's just going to be the index n squared. And this right here will be the nth partial sum, and then these are the first four partial sums, and that is the solution to our problem. So hope you found this video helpful, and let's move on.

Compute the first four partial sums and find a formula for the nth⁡$$n^{\operatorname{th}$$}nth partial sum.∑n=1∞2n−1\sum_{n=1}^{\infty}2n-1∑n=1∞2n−1

1,4,9,16;Sn=n21,4,9,16;S_{n}=$$n^21$$,4,9,16;Sn=n2

1,3,5,9;Sn=2n−11,3,5,9;S_{n}=2n-11,3,5,9;Sn=2n−1

1,4,9,16;Sn=2n−11,4,9,16;S_{n}=2n-11,4,9,16;Sn=2n−1

1,3,5,9;Sn=n21,3,5,9;S_{n}=$$n^21$$,3,5,9;Sn=n2

14. Sequences & Series

Series

Business Calculus

14. Sequences & Series

Series

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Multiple Choice

Compute the first four partial sums and find a formula for the $n^{$\operatorname{th}$}$ partial sum.
$$\sum$_{n=1}^{$\infty$}2n-1$

$1,3,5,9;S_{n}=2n-1$

$1,4,9,16;S_{n}=n^2$

$1,4,9,16;S_{n}=2n-1$

$1,3,5,9;S_{n}=n^2$

Verified step by step guidance

Step 1: Understand the problem. We are tasked with computing the first four partial sums of the series ∑_{n=1}^{∞}(2n−1) and finding a formula for the nth partial sum, S_n.

Step 2: Recall the definition of a partial sum. The nth partial sum, S_n, is the sum of the first n terms of the series. For this series, the terms are given by the expression (2n−1).

Step 3: Compute the first four terms of the series. Substitute n=1, n=2, n=3, and n=4 into the formula (2n−1) to find the terms: For n=1: 2(1)−1=1 For n=2: 2(2)−1=3 For n=3: 2(3)−1=5 For n=4: 2(4)−1=7.

Step 4: Compute the first four partial sums by adding the terms sequentially: S_1 = 1 S_2 = 1+3=4 S_3 = 1+3+5=9 S_4 = 1+3+5+7=16.

Step 5: Observe the pattern in the partial sums. The first four partial sums are 1, 4, 9, and 16. These correspond to perfect squares: 1^2, 2^2, 3^2, and 4^2. Thus, the formula for the nth partial sum is S_n = n^2.

9:06m

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Compute the first four partial sums and find a formula for the nthn^{\(\operatorname{th}\)}nth partial sum.∑n=1∞2n−1\(\sum\)_{n=1}^{\(\infty\)}2n-1∑n=1∞2n−1

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Compute the first four partial sums and find a formula for the $n^{\(\operatorname{th}\)}$ partial sum.
$\(\sum\)_{n=1}^{\(\infty\)}2n-1$