SUM, from i=1 to i=4, of (x_i - 3)^2; sum, from i=1 to i=4, of (x_i - 3)
- Thread starter Barbra
- Start date
Harry_the_cat
Elite Member
- Joined
- Mar 16, 2016
- Messages
- 3,693
The second one is correct.
The first one isn't. Note that \(\displaystyle (a-b)^2\) does NOT equal \(\displaystyle a^2-b^2\).
The first one isn't. Note that \(\displaystyle (a-b)^2\) does NOT equal \(\displaystyle a^2-b^2\).
Harry_the_cat
Elite Member
- Joined
- Mar 16, 2016
- Messages
- 3,693
one question per thread please.
Summations are not hard but you must be careful. The details can trip you up. Expand your terms before you break the sums up.
[math] \sum_{i=1}^4 (x_i - 3)^2 = \sum_{i=1}^4 (x_i^2 - 6x_i + 9) =\\ \left ( \sum_{i=1}^4 x_i^2 \right ) - 6 * \left ( \sum_{i=1}^4x_i \right ) + 9 * \left ( \sum_{i=1}^4 1 \right) = \text {WHAT?} [/math]
[math] \sum_{i=1}^4 (x_i - 3)^2 = \sum_{i=1}^4 (x_i^2 - 6x_i + 9) =\\ \left ( \sum_{i=1}^4 x_i^2 \right ) - 6 * \left ( \sum_{i=1}^4x_i \right ) + 9 * \left ( \sum_{i=1}^4 1 \right) = \text {WHAT?} [/math]
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,079
As I read these problems, you are given the values of the terms, so you don't need to do any expansion (indeed, any algebra at all); just plug in the numbers they gave, in the form in which the summation is given.
So, what are the four values of [imath](x_i-3)^2[/imath] that you need to sum?
Of course, by doing the expansion (incorrectly), you have show an error in your thinking, which gives us a chance to help.