seja f(n) a soma dos n termos de uma progressao aritmetica. demonstrar que f (n+3) -3f (n+2) +3f(n+1) - f(n) =0
Olá, Alberico.
[tex]f(n+3) -3f (n+2) +3f(n+1) - f(n)=[/tex]
[tex]=(n+3)\frac{a_1+a_{n+3}}{2}-3(n+2)\frac{a_1+a_{n+2}}{2}+3(n+1)\frac{a_1+a_{n+1}}{2}-[/tex]
[tex]n\cdot\frac{a_1+a_{n}}{2}=[/tex]
[tex]=(n+3)\frac{a_1+a_{n}+3r}{2}-3(n+2)\frac{a_1+a_{n}+2r}{2}+3(n+1)\frac{a_1+a_{n}+r}{2}[/tex]
[tex]-n\cdot\frac{a_1+a_{n}}{2}=[/tex]
[tex]=\frac12\cdot(na_1+na_n+3nr+3a_1+3a_n+9r-3na_1-3na_n-6nr-[/tex]
[tex]6a_1-6a_n-12r+3na_1+3na_n+3nr+3a_1+3a_n+3r-na_1-[/tex]
[tex]na_n)=[/tex]
[tex]=\frac12\cdot(na_1-3na_1+3na_1-na_1+na_n-3na_n+3na_n-na_n+[/tex]
[tex]3nr-6nr+3nr+3a_1-6a_1+3a_1+3a_n-6a_n+3a_n+9r-[/tex]
[tex]12r+3r)=[/tex]
[tex]=0[/tex]
