# Math Help - An expression for a sum

1. ## An expression for a sum

Hi, I am going through some calculus questions and came across this particular question which I don't really understand. It says:
Consider the sum
$\sum_{j=1}^{n}(3j-1)=2+5+8+11+...+3n-1$
We haven't spent a great deal of time on summations in lectures, so I was quite confused how to approach this, as isn't the sigma the expression for this sum... Any help would be greatly appreciated.

2. Originally Posted by macduff
Hi, I am going through some calculus questions and came across this particular question which I don't really understand. It says:
Consider the sum
$\sum_{j=1}^{n}(3j-1)=2+5+8+11+...+3n-1$
We haven't spent a great deal of time on summations in lectures, so I was quite confused how to approach this, as isn't the sigma the expression for this sum... Any help would be greatly appreciated.
Are you familiar with the summation notation? This might help:

In long hand the expression is

(3(1) - 1) + (3(2) - 1) + (3(3) - 1) + (3(4) - 1) + .... + (3(n) - 1)

= (3 - 1) + (6 - 1) + (9 - 1) + (12 - 1) + .... + (3n - 1)

= 2 + 5 + 8 + 11 + .... + 3n - 1

3. Sorry, I didn't post the end of the question, it was supposed to say find and expression for this sum. Thats why I am so confused because the sum is 2 + 5 + 8 + 11 + .... + 3n - 1 so isn't the expression of the sum the summation notation? Maybe the original question in the book has a typo in it...
Originally Posted by mr fantastic
Are you familiar with the summation notation? This might help:

In long hand the expression is

(3(1) - 1) + (3(2) - 1) + (3(3) - 1) + (3(4) - 1) + .... + (3(n) - 1)

= (3 - 1) + (6 - 1) + (9 - 1) + (12 - 1) + .... + (3n - 1)

= 2 + 5 + 8 + 11 + .... + 3n - 1

4. Originally Posted by macduff
Sorry, I didn't post the end of the question, it was supposed to say find and expression for this sum. Thats why I am so confused because the sum is 2 + 5 + 8 + 11 + .... + 3n - 1 so isn't the expression of the sum the summation notation? Maybe the original question in the book has a typo in it...
$\sum_{j=1}^n (3j - 1) = 3 \left( \sum_{j=1}^n j \right) - \sum_{j=1}^n 1 = 3 \cdot \frac{n(n+1)}{2} - n$.

The simplification of this answer is left for you to do.

The research of the formula I used is left for you to do eg. How to Sum the Integers from 1 to N - wikiHow

5. Thanks!
Originally Posted by mr fantastic
$\sum_{j=1}^n (3j - 1) = 3 \left( \sum_{j=1}^n j \right) - \sum_{j=1}^n 1 = 3 \cdot \frac{n(n+1)}{2} - n$.

The simplification of this answer is left for you to do.

The research of the formula I used is left for you to do eg. How to Sum the Integers from 1 to N - wikiHow