Question #c18f2

1 Answer
NickTheTurtle
Apr 13, 2017

#7# terms in total:
#27+9+3+1+1/3+1/9+1/27=1093/27#

Explanation:

The series is a geometric series and is thus in the form #a_n=a_(n-1)*r# or #a_n=a_1*r^(n-1)#.

The first term, #a_1#, is #27#. The common ratio #r# is #1/3#.

We need to find #n# such that #sum_(k=1)^n27*(1/3)^(k-1)=1093/27#.

The sum of a geometric series is equal to #(a_1(1-r^(n-1)))/(1-r)#. Thus, the left-hand side can be expressed as #(27(1-(1/3)^(n-1)))/(1-1/3)=81/2(1-1/3^(n-1))#.

Solve #81/2(1-1/3^(n-1))=1093/27#. This becomes #1/3^(n-1)=1-2186/2187=1/2187#. Solving for #n# results in #n=7#.

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