Question

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

Answer 1

Olá, Alberico.

 

 

$<var>f(n+3) -3f (n+2) +3f(n+1) - f(n)=</var>$

 

 

$=<var>(n+3)\frac{a_1+a_{n+3}}{2}-3(n+2)\frac{a_1+a_{n+2}}{2}+3(n+1)\frac{a_1+a_{n+1}}{2}-$

$n\cdot\frac{a_1+a_{n}}{2}=</var>$

 

 

$=<var>(n+3)\frac{a_1+a_{n}+3r}{2}-3(n+2)\frac{a_1+a_{n}+2r}{2}+3(n+1)\frac{a_1+a_{n}+r}{2}</var>$

$<var>-n\cdot\frac{a_1+a_{n}}{2}=</var>$

 

 

$<var>=\frac12\cdot(na_1+na_n+3nr+3a_1+3a_n+9r-3na_1-3na_n-6nr-</var>$

$<var>6a_1-6a_n-12r+3na_1+3na_n+3nr+3a_1+3a_n+3r-na_1-$

$na_n)=</var>$

 

 

$<var>=\frac12\cdot(na_1-3na_1+3na_1-na_1+na_n-3na_n+3na_n-na_n+$

$3nr-6nr+3nr+3a_1-6a_1+3a_1+3a_n-6a_n+3a_n+9r-</var><var>$

$<var>12</var><var>r+</var>3r)=</var>$

 

 

$<var>=0</var>$