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Question

Mathematics Question on Sequences and Series

Find the sum of the following series up to n terms: (i) 5+55+555+…(ii) .6+.66 +. 666+…

Answer

(i) 5 + 55 + 555 + …
Let Sn = 5 + 55 + 555 + ….. to n terms
=59[9+99+999+.....to  n  terms]= \frac{5 }{ 9} [9 + 99 + 999 + ..... to\,\, n\,\, terms]
=59[(10−1)+(102−1)+(103−1)+....to  n  terms]= \frac{5 }{ 9} [(10 - 1) + (102 - 1) + (103 - 1) + ....to\,\, n\,\, terms]
=59[(10+102+103+......to   n   terms)−(1+1+....n  terms)]=\frac{ 5 }{ 9} [(10 + 102 + 103 + ...... to\,\,\, n\,\,\, terms) - (1 + 1+ ....n \,\,terms)]
=59[10(10n−1)10−1−n]= \frac{5 }{ 9} [\frac{10 (10n - 1) }{10 - 1} - n]
=59[10(10n−1)9−n]=\frac{ 5 }{ 9} [\frac{10 (10n - 1) }{ 9 }- n]
=5081(10n−1)−5n9= \frac{50 }{ 81 }(10n - 1) -\frac{ 5n }{9}
(ii) .6 +.66 +. 666 +…
Let Sn = 06. + 0.66 + 0.666 + … to n terms
= 6 [0.1 + 0.11 + 0.111 + .... to n terms]
=69\frac{ 6 }{9}[0.9 + 0.99 + 0.999 + ... to n terms]
= 69[(1−110)+(1−1102)+(1−1103)+\frac{6 }{ 9} [(1 - \frac{1 }{ 10}) + (1 - \frac{1 }{ 10^2}) + (1 - \frac{1 }{ 10^3}) + ... to n terms]
=23[(1+1+= \frac{2 }{ 3} [(1 + 1 + .... n terms)$$ - \frac{1 }{ 10 }1+(\frac{1 }{10} + \frac{1 }{10^2} + .... n terms)]
=23[n−110(1−(110)21−110)]=\frac{ 2 }{ 3} [ n -\frac{ 1 }{ 10} (\frac{1-(\frac{1}{10})^2}{1-\frac{1}{10}})]
=23n−230×109(1−10−n)= \frac{2 }{ 3} n -\frac{ 2 }{ 30} × \frac{10 }{ 9} (1 - 10 ^{- n})
=23n−227(1−10−n)=\frac{ 2 }{ 3} n - \frac{2 }{ 27} (1- 10 ^{- n})