Question

Sn=5+55+555+.....55......5(n cifre) , n ∈ N*

Answer 1

   
[tex]Sn=5+55+555+\hdots + 55 \hdots 5 \\ \texttt{Sirul are n termeni, iar ultimul termen are n cifre. } \\ \texttt{Vom face transformari astfel incat in loc de } \\ \texttt{cifra 5 sa avem cifra 9 } \\ \\ 5+55+555+\hdots + 55 \hdots 5 = \\ = 5(1+11+111+\hdots + 11 \hdots 1) = \\ \\ =5\times \frac{9}{9} \times (1+11+111+\hdots + 11 \hdots 1) = \\ \\ = \frac{5}{9} (9+99+999+\hdots + 99 \hdots 9) = ~\text{Acum adunam 1 la fiecare termen.}[/tex]


[tex]=\frac{5}{9} (9+99+999+\hdots + 99 \hdots 9 + n - n) = \\ \\ =\frac{5}{9} (9+1+99+1+999+1+\hdots + 99 \hdots 9+1 - n) = \\ \\ =\frac{5}{9} (10+100+1000+\hdots + 10 \hdots 0 - n) = \\ \\ \texttt{Avem n termeni din care ultimul are 1 urmat de n zerouri.} \\ \texttt{Ptimul termen are 2 cifre si termenul n are n+1 cifre.} \\ \\ = \frac{5}{9}(111111....de ~n ~ori ....110 -n) = \displaystyle \\ \\ \texttt{Nr. are n de 1 si un zero.} \\ \\ = \frac{5((111111....de ~n ~ori ....110 -n) )}{9} =[/tex]


[tex]\displaystyle \\ = \boxed{\frac{555555....de ~n ~ori ....550 -5n}{9} } \\ \\ \texttt{La numarator avem un numar format din n de 5 urmat de 0,} \\ \texttt{din care se scade 5n, iar la numitor este 9.}[/tex]