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Assuming that $\bar{x} = \frac{\sum_{i = 1}^n x_i}{n}$, I have the following expression that I'm trying to simplify:

$\sum_{i = 1}^n (x_i - \bar{x})^2 + n\bar{x}^2$

$ = \sum_{i = 1}^n (x_i^2 - 2\bar{x}x_i + \bar{x}^2) + n\bar{x}^2$

$ = \sum_{i = 1}^n (x_i^2) - 2n\bar{x}\sum_{i = 1}^n x_i + n\bar{x}^2 + n\bar{x}^2$

$ = \sum_{i = 1}^n (x_i^2) - 2(\sum_{i = 1}^n x_i)^2 + \frac{2(\sum_{i = 1}^n x_i)^2}{n}$

However, the solution is

$ = \sum_{i = 1}^n x_i^2$?

How did $-2(\sum_{i = 1}^n x_i)^2 + \frac{2(\sum_{i = 1}^n x_i)^2}{n} = 0$?

Adrian
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  • You incorrectly equate $\sum 2\overline x x_i$ with $2n\overline x\sum x_i$. (the $n$ shouldn't be there). – lulu Sep 19 '16 at 19:17
  • So what's the correct expression then? $2\bar{x}\sum_{i = 1}^n x_i$? I thought we're summing $\bar{x}$ $n$ times? – Adrian Sep 19 '16 at 19:18
  • Same as you wrote, but without the $n$. The $2\overline x$ is just a constant...it simply factors out of the sum. – lulu Sep 19 '16 at 19:19
  • Clear now? if $\lambda$ is a constant then $\sum \lambda x_i=\lambda \sum x_i$. That's just the distributive law for addition. You handle the last term correctly, as $\sum \overline x=\overline x\sum 1=n\overline x$. – lulu Sep 19 '16 at 19:28

1 Answers1

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As lulu commented, your mistake was made at the second equal sign. It should be: $$=\sum_{i=1}^n x_i^2 - 2\bar x \sum_{i=1}^n x_i + n\bar{x}^2 + n\bar{x}^2 = \sum_{i=1}^n x_i^2 -2\bar{x}(n\bar{x}) + 2n\bar{x}^2 = \sum_{i=1}^n x_i^2.$$

trang1618
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