Ensino Médio ⇒ progressão Tópico resolvido

[bdisvan]

Determine a soma dos n primeiros termos da sequência:
S=(5,55,555,5555...)

[petras]

[tex3]\mathsf{

S_n =5+55+555+... n \\
\therefore S_n =5(1+11+111+... n )\\
⇒S_n =\frac{5}{9} (9+99+999+... n)\\
⇒S_n =\frac{5}{9} [{(10−1)+(10^2 −1)+(10^3 −1)+...+(10^n −1)}]\\
⇒S_n =\frac{5}{9} [{(\underbrace{10+10^2+10^3 +...+10^n }_{S_{PG}:a_1=10:q=10})−n}]\\
⇒S_ n =\frac{5}{9} [{10[\frac{ 10^n−1}{10 −1 }]−n}]\\
⇒S_n =\frac{5}{9} [\frac{10}{9} (10^n −1)−n]=\boxed{\frac{50}{81}( 10^{n-1}) −\frac{5n}{9})}\color{green}\checkmark
}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{

S_n =5+55+555+... n \\
\therefore S_n =5(1+11+111+... n )\\
⇒S_n =\frac{5}{9} (9+99+999+... n)\\
⇒S_n =\frac{5}{9} [{(10−1)+(10^2 −1)+(10^3 −1)+...+(10^n −1)}]\\
⇒S_n =\frac{5}{9} [{(\underbrace{10+10^2+10^3 +...+10^n }_{S_{PG}:a_1=10:q=10})−n}]\\
⇒S_ n =\frac{5}{9} [{10[\frac{ 10^n−1}{10 −1 }]−n}]\\
⇒S_n =\frac{5}{9} [\frac{10}{9} (10^n −1)−n]=\boxed{\frac{50}{81}( 10^{n-1}) −\frac{5n}{9})}\color{green}\checkmark
}[/tex3]

(Solução: net)