# Thread: Partial Sums of a Series

1. ## Partial Sums of a Series

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:

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

$
=\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?

2. 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:

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

$
=\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?

It would actually be

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

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

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

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

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

3. 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.

4. 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 $3\left(\sum_{i = 1}^n{i}\right) = \frac{3n}{2}(1 + n)$.

5. 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!