Partial Sums of a Series

**Kasper** · September 27th 2009, 08:47 AM

Hey folks, been an awful long time since I've been working with series, could definitely use a push with this problem.

The series obtained from the sequence $a_{n}=3n+2$ , is the restriction to the integers of a polynomial function $y=f(x)$ . Construct the partial sums of such series and use them to identify the function $f(x)$ .

I started by writing the series in sigma notation,

$\sum_{i=1}^n (3i+2)$

And to construct the partial sums, we can split up the total sum into:

$\sum_{i=1}^n (3i+2)$

$<br /> =\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)$

$=3 \sum_{i=1}^n (i) + 2$

Right?

I want to say that $f(x)=3x+2$ which makes sense as it would be $a_{n}=3n+2$ if restricted to the integers, but I'm not sure if I've found my correct partial sum?

Thanks for your time!

**Prove It** · September 27th 2009, 08:55 AM

Originally Posted by Kasper

Hey folks, been an awful long time since I've been working with series, could definitely use a push with this problem.

The series obtained from the sequence $a_{n}=3n+2$ , is the restriction to the integers of a polynomial function $y=f(x)$ . Construct the partial sums of such series and use them to identify the function $f(x)$ .

I started by writing the series in sigma notation,

$\sum_{i=1}^n (3i+2)$

And to construct the partial sums, we can split up the total sum into:

$\sum_{i=1}^n (3i+2)$

$<br /> =\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)$

$=3 \sum_{i=1}^n (i) + 2$

Right?

I want to say that $f(x)=3x+2$ which makes sense as it would be $a_{n}=3n+2$ if restricted to the integers, but I'm not sure if I've found my correct partial sum?

Thanks for your time!

It would actually be

$=\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)$

$= 3\left(\sum_{i = 1}^n {i}\right) + 2n$

$= \frac{3n}{2}(1 + n) + 2n$

$= \frac{3n}{2} + \frac{3n^2}{2} + 2n$

$= \frac{3n^2}{2} + \frac{7n}{2}$ .

**Kasper** · September 27th 2009, 09:00 AM

Originally Posted by Prove It

It would actually be

$=\sum_{i=1}^n (3i) + \sum_{i=1}^n (2)$

$= 3\sum_{i = 1}^n (i) + 2n$

$= \frac{3}{2}(1 + n) + 2n$

$= \frac{3}{2} + \frac{3}{2}n + 2n$

$= \frac{3}{2} + \frac{7}{2}n$ .

Ah, I missed how the sum to n terms of 2 would be 2n, of course. :/

I am however still confused on how $\sum_{i=1}^n (3i) = \frac{1}{2}(1+n)$ . Series have definitely never been a strong point for me.

**Prove It** · September 27th 2009, 09:07 AM

Originally Posted by Kasper

Ah, I missed how the sum to n terms of 2 would be 2n, of course. :/

I am however still confused on how $\sum_{i=1}^n (3i) = \frac{1}{2}(1+n)$ . Series have definitely never been a strong point for me.

Note, I edited the above post.

We have

$\sum_{i = 1}^n {(3i)} = 3\left(\sum_{i = 1}^n {i}\right)$

Notice that $\sum_{i = 1}^n {i} = 1 + 2 + 3 + \dots + n$ .

This is an arithmetic series with $t_1 = 1$ , $t_n = n$ , and common difference $d = 1$ .

The sum of an arithmetic sequence can be found using the formula

$\sum_{i = 1}^n{i} = \frac{n}{2}(t_1 + t_n)$

$= \frac{n}{2}(1 + n)$ .

Thus $3\left(\sum_{i = 1}^n{i}\right) = \frac{3n}{2}(1 + n)$ .

**Kasper** · September 27th 2009, 09:14 AM

Originally Posted by Prove It

Note, I edited the above post.

We have

$\sum_{i = 1}^n {(3i)} = 3\left(\sum_{i = 1}^n {i}\right)$

Notice that $\sum_{i = 1}^n {i} = 1 + 2 + 3 + \dots + n$ .

This is an arithmetic series with $t_1 = 1$ , $t_n = n$ , and common difference $d = 1$ .

The sum of an arithmetic sequence can be found using the formula

$\sum_{i = 1}^n{i} = \frac{n}{2}(t_1 + t_n)$

$= \frac{n}{2}(1 + n)$ .

Thus $3\left(\sum_{i = 1}^n{i}\right) = \frac{3n}{2}(1 + n)$ .

Oh! That makes perfect sense, I'll have to keep that formula handy, thanks so much!

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