tex]...

July 2022 1 15 Report

Найдите натуральное число n, для которого выполняется равенство
(c объяснением):
[tex] \frac{4 ^{3}-1 }{4 ^{3}+1 } * \frac{5 ^{3}-1 }{5 ^{3}+1 } *...* \frac{ n^{3}-1 }{n^{3}+1 } = \frac{3906}{4225} [/tex]

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dtnth

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$A^3-B^3=(A-B)(A^2+AB+B^2)$
$A^3+B^3=(A+B)(A^2-AB+B^2)$
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$\frac{4^3-1}{4^3+1}*\frac{5^3-1}{5^3+1}*...*\frac{n^3-1}{n^3+1}=$
$\frac{(4-1)*(4^2+4*1+1^2)*(5-1)*(5^2+5*1+1)*....*(n-1)(n^2+n*1+1^2)}{(4+1)*(4^2-4*1+1^2)*(5+1)*(5^2-5*1+1)*....*(n+1)(n^2-n*1+1^2)}=$
$\frac{(4-1)*(5-1)}{(n-1)+1)(n+1}*\frac{(6-1)*(7-1)*...*(n-1)}{(4+1)*(5+1)*....((n-2)+1)}$ *
$*\frac{(4*(4+1)+1)(5*(5+1)+1)*(6*(6+1)+1)*...*(n(n+1)+1)}{4*(4-1)+1)*(5*(5-1)+1)*...*(n(n-1)+1)}=$
$\frac{3*4}{n(n+1)}*$
$*\frac{(4*5+1)(5*6+1)*(6*7+1)*(7*8+1)*...*(n(n+1)+1)}{(3*4+1)(4*5+1)*(5*6+1)*(6*7+1)*....*((n-1)n+1)}=$
$\frac{3*4}{n(n+1)}*\frac{n(n+1)+1}{3*4+1}*\frac{(4*5+1)*(5*6+1)*,,,*((n-1)n+1)}{(4*5+1)*(5*6+1)*...*((n-1)n+1)}=$
$\frac{12(n^2+n+1)}{13(n^2+n)}$
===========================
$\frac{12(n^2+n+1)}{13(n^2+n)}=\frac{3906}{4225}$
$\frac{n^2+n+1}{n^2+n}=\frac{3906*13}{4225*12}$
$1+\frac{1}{n^2+n}=\frac{651}{325*2}$
$1+\frac{1}{n^2+n}=\frac{651}{650}=\frac{650+1}{650}=1+\frac{1}{650}$
$\frac{1}{n^2+n}=\frac{1}{650}$
$n^2+n=650$
$n^2+n-650=0$
$D=1^2-4*1*(-650)=2601=51^2$
$n_1=\frac{-1-51}{2*1}$ - не подходит (n должно быть натуральным)
$n_2=\frac{-1+51}{2*1}=25$
$n=25$
ответ: 25

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Vlаdislav Спасибо

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