sum of n terms of the series 0.1+0.11+0.111+...is
1/9{ n-(1 - ( 0.1 ) ⁿ) }
n-1 - (0.1)ⁿ /9
1/9 {n- (1-(0.1) ⁿ} /9
pls solve this....
ROSHAN PAUL JOSEPH
(CA Final student)
(311 Points)
Replied 02 June 2011
[0.1+0.11+0.111+………] *9/9
=[0.9+0.99+0.999+…….]*1/9
=[(1-0.1)+ (1-0.01)+ (1-0.001)+….]*1/9
=n-(0.1+0.01+0.001+…….) * 1/9
(0.1+0.01+0.001+…….) is a G.P with common ratio 0.1;
Sum of terms of G.P = a(1-rn)/(1-r)
i.e, (0.1+0.01+0.001+…….) = 0.1(1-0.1n)/(1-0.1)
i.e,( 1-0.1n )/9
therefore it becomes , {n-( 1-0.1n )/9}*1/9
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