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

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