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Middle Terms

if 1+x+1/x4=∑...

Question

if $(1 + x + \frac{1}{x})^{4} = \sum_{r_{1} + r_{2} + r_{3}} = 4 \frac{4!}{r_{1}! \times r_{2}! \times r_{3}!} \times (1)^{r_{1}} (x)^{r_{2}} (\frac{1}{x})^{r_{3}}$ how many values can $r_{1}$ take so that the terms in the expansion will be independent of x.

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Solution

We want the terms to be independent of x.
$\Rightarrow r_{2} = r_{3}$
So the values $(r_{1}, r_{2}, r_{3})$ can take are (4,0,0),(2,1,1) and (0,2,2). This is because we want $r_{1} + r_{2} + r_{3} = 4$ and $r_{2} = r_{3}$ .
$\Rightarrow r_{1}$ can take three values.

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\frac{R_{1}}{r_{1}} + \frac{R_{2}}{r_{2}} + \frac{R_{3}}{r_{3}} =

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r_{1} + r_{2} + r_{3} - r = 4 R

r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = Δ^{2}

△ A B C

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r < r_{1} < r_{2} < r_{3}

△ A B C, r_{1} r_{2} + r_{2} r_{3} + r_{3} r_{1} = \frac{r_{1} r_{2} r_{3}}{r}

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