The sum of 100 terms of the series 0.9+0.09+0.009+.... will be?

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Answer 1 · 2018-04-28T14:38:36

Sum #approx 1#

Sum #= 0.9+0.09+0.009+ ...#

The sum is a geometric progression (GP) with first term #(a_1)=0.9# and common ratio #(r)=0.1#

The sum of the first #n# terms of a GP is given by:

Sum #= (a_1(1-r^n))/(1-r)#

In this case,

Sum #= (0.9(1-0.1^100))/(1-0.1)#

#= (cancel0.9(1-0.1^100))/cancel0.9#

#approx (1-0)#

#approx 1#

[NB: #0.1^100 = 1 xx 10^-100# so #(1- 0.1^100)# is actually #0.# followed by 100 #9#'s]