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For $i = 1,2,\ldots,19$ we have:

$$\left(\sum_{j=1, j\neq i}^{19}\alpha_j\right)\cdot(1.5) + (\alpha_i)\cdot(3) = 1.5 \implies \left(\sum_{j=1}^{19}\alpha_j\right)(1.5) + (\alpha_i)(1.5) = 1.5$$

Does anyone know how to prove this? Any intuition behind it?

I thought it would make sense when written as:

$$\alpha_1 = \frac 1 2 - \frac 1 2 \left(\sum_{j=2}^{19}\alpha_j\right)$$

$$ \vdots $$

$$\alpha_{19} = \frac 1 2 - \frac 1 2 \left( \sum_{j=1}^{18}\alpha_j \right)$$

$\alpha_i$ are unknowns

But I still fail to see why.

Michael Hardy
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Mahamad A. Kanouté
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    Is, say, $\alpha_j(1.5)$ multiplication or function evaluation? – Ian Nov 27 '20 at 01:28
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    Your notation is hard to follow. First of all, writing $\sum a_i+b$ is ambiguous. Do you mean $\left(\sum a_i\right)+b$ or $\sum (a_i+b)$? Secondly, what does $a_i(n)$ mean? – lulu Nov 27 '20 at 01:31

1 Answers1

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The desired claim follows quickly from the identity: $$\left(\sum_{j=1,\,j\neq i}^{19}a_j\right)+a_i=\sum_{j=1}^{19}a_j$$

Now just multiply both sides by $1.5$ and add $1.5\times a_i$ to both sides.

lulu
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