求下列数列的前n项和Sn：5，55，555，5555，…，59(10n−1)，…．

Question

_n(5)/(9)(10n−1)，…．(5)/(9)55990n−1)，…．0n−1)，…．n−1)，…．

Answer 1

Sn＝5+55+555+…+

n个

5…5

=(5)/(9)(9+99+999++

n个

999

)
=(5)/(9)[(10−1)+(102−1)+(103−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n Sn＝5+55+555+…+

n个

5…5

=(5)/(9)(9+99+999++

n个

999

)
=(5)/(9)[(10−1)+(102−1)+(103−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n n＝5+55+555+…+

n个

5…5

n个n个n个5…55…55…5
=(5)/(9)(9+99+999++

n个

999

)
=(5)/(9)[(10−1)+(102−1)+(103−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n (5)/(9)555999(9+99+999++

n个

999

n个n个n个

999

999

999

999999999)
=(5)/(9)[(10−1)+(102−1)+(103−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n (5)/(9)555999[(10−1)+(102−1)+(103−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n 2−1)+(103−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n 3−1)++(10n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n n−1)]
=(5)/(9)[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n (5)/(9)555999[10+102+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n 2+103++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n 3++10n−n]＝(50)/(81)(10n−1)−(5)/(9)n n−n]＝(50)/(81)505050818181(10n−1)−(5)/(9)n n−1)−(5)/(9)555999n