# Evaluate the sum

1. Dec 15, 2012

### sharks

1. The problem statement, all variables and given/known data
$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})]$$

2. Relevant equations
Multiple-sigma notation.

3. The attempt at a solution
I agree this seems like basic summation stuff, but i do not agree with the given answer. So, here is what i've done.

$$\sum_{i=1}^{m} \sum_{j=1}^{n}[2(x_i-x_{i-1})-3(y_j-y_{j-1})] \\=2\sum_{i=1}^{m} (x_i-x_{i-1})-3\sum_{j=1}^{n}(y_j-y_{j-1}) \\=2(x_m-x_0)-3(y_n-y_0)$$

However, the given answer for this problem is:
$$2n(x_m-x_0)-3m(y_n-y_0)$$

2. Dec 15, 2012

### Zondrina

Do you know what the partition of the interval is? Was there an interval given in the question?

3. Dec 15, 2012

### Dick

The terms that are indexed with i are independent of j. So when you apply the j sum they will get added to themselves n times. So they will pick up a factor of n. Similarly for the terms indexed with j.

4. Dec 15, 2012

### Staff: Mentor

You're skipping some steps above.

After the first line you should have this:
$$\sum_{i=1}^{m} \left(\sum_{j=1}^{n}[2(x_i-x_{i-1}] - \sum_{j=1}^{n}[3(y_j-y_{j-1})]\right)$$
The first sum isn't affected by j, so will simplify to n * the expression being summed. Something similar happens for the outer summation.

5. Dec 15, 2012

### sharks

Yes, there are two intervals given, but in this case, the answer still wouldn't be correct, so i omitted that part from the stated problem in the 1st post. Anyway, here they are:

Let $P_1 = \{ x_0, x_1, x_2,..., x_m \}$ be a partition of $[a_1, a_2]$
Let $P_2 = \{ y_0, y_1, y_2,..., y_n \}$ be a partition of $[b_1, b_2]$

That just cleared up all my confusion! Clear and concise explanation. Thanks, Dick!

I had intentionally skipped those steps, as all the terms in i are independent of j, and all the terms in j are independent of i. So, in the expansion, i had wrongly assumed that expanding all the i terms with respect to j and all the j terms with respect to i, would give 1. Thanks for the help.