Question

Решите уравнение:
$5 C^{n-1}_{2n}=8 C^{n}_{2n-1}$

Answer 1

$5C_{2n}^{~n-1}=8C_{2n-1}^{~n}\\dfrac{5cdot (2n)!}{(2n-n+1)!cdot(n-1)!}=dfrac{8cdot (2n-1)!}{(2n-1-n)!cdot n!}\\dfrac{5cdot (2n-1)!cdot 2n}{(n+1)cdot n!cdot(n-1)!}=dfrac{8cdot (2n-1)!}{(n-1)!cdot n!}~~~~~Bigg|cdot dfrac{(n-1)!cdot n!}{(2n-1)!}\\dfrac{10n}{(n+1)}=dfrac{8}1\\10n=8n+8\2n=8\boxed{boldsymbol{n=4}}$

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Проверка :

$5C_{2n}^{~n-1}=8C_{2n-1}^{~n}~;~~~~n=4\\5C_8^3=8C_7^4\\dfrac{5cdot 8!}{(8-3)!cdot3!}=dfrac{8cdot 7!}{(7-4)!cdot 4!}\\dfrac{5cdot 5!cdot 6cdot 7cdot 8}{5!cdot2cdot 3}=dfrac{8cdot 4!cdot 5cdot 6cdot 7}{2cdot 3cdot 4!}\\5cdot 7cdot 8=8cdot 5cdot 7$