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I want to solve for the constant $c$

$\sum_{x_2 = 0}^{2} \sum_{x_1 = 0 }^{2} c(x_1 + x_2) = 1$

So I expanded out the sum

$\sum_{x_2 = 0}^{2} \sum_{x_1 = 0 }^{2} c(x_1 + x_2) = \sum_{x_2 = 0}^{2} \sum_{x_1 = 0 }^{2} cx_1 + cx_2 = \sum_{x_2 = 0}^{2} (\sum_{x_1 = 0 }^{2} cx_1 + \sum_{x_1 = 0 }^{2} cx_2) = \sum_{x_2 = 0}^{2}(3c + 2cx_2) = 3c\sum_{x_2 = 0}^{2}1 + 2c\sum_{x_2 = 0}^{2}x_2 = 3c(2) + 2c(3) = 12c = 1 \iff c = 1/12$

According to Mathematica, the constant should be 1/18.

I cannot figure out what I did wrong for the life of me

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3 Answers 3

up vote 4 down vote accepted

In this step, $$\sum_{x_2 = 0}^2\left(\sum_{x_1 = 0}^2cx_1 + \sum_{x_1 = 0}^2cx_2\right)$$ you made a mistake with the latter sum. It should evaluate to $$\sum_{x_2 = 0}^2\left(3c + 3cx_2\right)$$ because in the latter sum you are summing over $3$ terms of $x_1$. Similarly, $$3c\sum_{x_2=0}^21 = 9c$$ because again, you are summing over three terms $\{0,\ 1,\ 2\}$.

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Twice you replaced the three-fold sum over a constant by $2$ times the constant instead of $3$ times the constant. The sum from $0$ to $2$ has three terms. If you fix that, you get the right answer.

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$\sum_{x_1=0}^2cx_2=3cx_2$, not $2cx_2$: $x_1$ runs through three values, $0,1$, and $2$, not just $2$ values.

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