[tex]a_1=\frac{k}{6} \\\\a_2=\frac{k}{12} \\\\a_3=\frac{k}{20} \\\\...\\\\a_n_-_1=\frac{k}{n(n+1)} \\\\a_{n}=\frac{k}{(n+1)(n+2)}[/tex]
[tex]\frac{k}{6} +\frac{k}{12} +\frac{k}{20} +...+\frac{k}{n(n+1)}+\frac{k}{(n+1)(n+2)} =5760\\\\k(\frac{1}{6} +\frac{1}{12} +\frac{1}{20} +...+\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)} )=5760\\\\k(\frac{1}{2}-\frac{1}{3} +\frac{1}{3}-\frac{1}{4} +\frac{1}{4}-\frac{1}{5} +...+\frac{1}{n}-\frac{1}{n+1} +\frac{1}{n+1}-\frac{1}{n+2} )=5760[/tex]
[tex]k(\frac{1}{2}- \frac{1}{n+2} )=5760[/tex]
[tex]k\times\frac{n}{2(n+2)} =5760[/tex]
[tex]a_{n-1}-a_n=\frac{k}{n(n+1) }-\frac{k}{(n+1)(n+2)} =40\\\\k(\frac{1}{n} -\frac{2}{n+1}+\frac{1}{n+2} )=40[/tex]
Numitorul comun este n(n+1)(n+2)
[tex]k\times\frac{n^2+3n+2-2n^2-4n+n^2+n}{n(n+1)(n+2)} =40\\\\k\times\frac{2}{n(n+1)(n+2)} =40\\[/tex]
[tex]k\times\frac{n}{2(n+2)} =5760[/tex]
impartim cele 2 relatii, a doua la prima si obtinem
[tex]\frac{n}{2(n+2)} \times \frac{n(n+1)(n+2)}{2} =144[/tex]
n²(n+1)=144×4
n³+n²-576=0
n³-512+n²-64=0
n³-8³+n²-8²=0
(n-8)(n²+8n+64)+(n-8)(n+8)=0
(n-8)(n²+9n+72)=0
n=8
[tex]k\times\frac{n}{2(n+2)} =5760[/tex]
[tex]k\times\frac{4}{10} =5760[/tex]
k=14 400
[tex]a_1=\frac{14400}{6} =2400[/tex]
[tex]a_n=\frac{14400}{9\times 10} =160[/tex]
