binomial help please (1 Viewer)

Drongoski · Sep 14, 2022

$= \left(px^3-qx^{-2} \right)^9 = \sum^9 _{i = 0} \binom 9 i (px^3)^{9-i} (qx^{-2})^i = \sum ^9 _{i=0}\binom 9 i p^{9-i} q^i x^{27-3i-2i}\\ \\ x^{17} $ term $ \implies 27 - 5i = 17 \implies i = 2 $ \\ \\ and $ x^{27} $ term $ \implies 27 - 5i = 27 \implies i = 0\\ \\ \therefore \binom 9 0 p^9 = \binom 9 2 p^7 q^2 \implies p^9 = 36p^7 q^2 \implies p^2 = 36q^2 \\ \\ \therefore p^2 - (6q)^2 = 0 \implies p^7(p+6q)(p-6q) = 0 \\ \\ \implies (p-6q) = 0 $ as $ p^7 $ and $ (p+6q) \neq 0$

QED

Masaken · Sep 14, 2022

Find the coefficients with the Tr+1 formula, I think coeff of x^17 is 36p^7q^2 and coeff of 27 is p^9. Equate them, move p^9 to the side of 36pq (blah, sorry not bothered to add powers), then factor out p^7, which will leave (36q^2 - p^2), which can be factored out with difference of two squares --> p^7(6q+p)(6q-p) = 0, then p^7 fades out of existence because p has to be a positive integer

I think from there you can apply null factor law and say 6q - p = 0, which can be rearranged to say p - 6q = 0

(Someone please correct if I am wrong, kind of uncertain about the last bit)

Edit: 2nd para is wrong, while 6q - p can be rearranged to say p - 6q is zero, you can't take the other factor as they both have to be positive integers (as the working out above me stated)

chilli 412 · Sep 14, 2022

sorry if this is messy

binomial help please (1 Viewer)

jkdC

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Drongoski

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Masaken

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chilli 412

oo la la

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