\
"Recall the result (formula) for the sum of a finite geometric"
"series:"
\quad sum_{k = 0}^n \ r^n \ = \ { r^{ n + 1 } - 1 } / { r - 1 }; \qquad \qquad \qquad "where" \ r \ "is the common ratio".
"The sum you have fits exactly this, with some small"
"adjustments."
"Notice that the fundamental formula above starts with the"
"index variable" \ [ k ] \ \ "at "0," and your sum starts with the"
"index variable" \ [ n ] \ \ "at "1."
"So let's start with your sum, and make the small adjustments"
"so we can use the fundamental formula above:"
\qquad \qquad \qquad sum_{ n = 1 }^9 \ 2^{ n - 1 } \ = \ [ 2^{ 0 - 1 } \ + \ sum_{ n = 1 }^9 \ 2^{ n - 1 } ] \ - \ 2^{ 0 - 1 }.
"Continuing, absorb the" \ \ 2^{ 0 - 1 } \ \ "term into the larger sum, since:"
2^{ 0 - 1 } = \ 2^{ n - 1 }, \ "with" \ n = 0:
\qquad \qquad \qquad \qquad \qquad = \ [ sum_{ n = 0 }^9 \ 2^{ n - 1 } ] \ - \ 2^{ - 1 }
\qquad \qquad \qquad \qquad \qquad = \ [ sum_{ n = 0 }^9 \ 2^{ - 1 }\cdot 2^{ n } ] \ - \ 2^{ - 1 }.
"Continuing, pull out the" \ 2^{ - 1 } \ "factor inside the last sum:"
\qquad \qquad \qquad \qquad \qquad = \ ( 2^{ - 1 } \cdot [ sum_{ n = 0 }^9 \ 2^{ n } ] ) \ - \ 2^{ - 1 }.
"Now pull out the" \ 2^{ - 1 } \ "factor from the two terms here:"
\qquad \qquad \qquad \qquad \qquad = \ 2^{ - 1 } \cdot ( [ sum_{ n = 0 }^9 \ 2^{ n } ] \ - \ 1 ).
"Thus we have now:"
\qquad \qquad \qquad sum_{ n = 1 }^9 \ 2^{ n - 1 } \ = \ 2^{ - 1 } \cdot ( [ sum_{ n = 0 }^9 \ 2^{ n } ] \ - \ 1 ).
"Now the sum inside the brackets fits exactly the"
"fundamental formula above -- using:" \ \ r = 2, n = 9,"in that formula. So, continuing:"
\qquad \qquad \qquad \qquad \qquad = \ 2^{ - 1 } \cdot ( [ { 2^{ 9 + 1 } - 1 } / { 2 - 1 } ] \ - \ 1 )
\qquad \qquad \qquad \qquad \qquad = \ 2^{ - 1 } \cdot ( [ (2^{ 10 } - 1 } / { 1 } ] \ - \ 1 )
\qquad \qquad \qquad \qquad \qquad = \ 2^{ - 1 } \cdot ( (2^{ 10 } - 1 ) \ - \ 1 )
\qquad \qquad \qquad \qquad \qquad = \ 2^{ - 1 } \cdot ( 2^{ 10 } - 1 \ - \ 1 )
\qquad \qquad \qquad \qquad \qquad = \ 2^{ - 1 } \cdot ( 2^{ 10 } - 2 )
\qquad \qquad \qquad \qquad \qquad = \ ( 2^{ - 1 } \cdot 2^{ 10 } ) - ( 2^{ - 1 } \cdot 2 )
\qquad \qquad \qquad \qquad \qquad = \ 2^{ 10 - 1 } - 2^{ 1 - 1 }
\qquad \qquad \qquad \qquad \qquad = \ 2^{ 9 } - 2^{ 0 }
\qquad \qquad \qquad \qquad \qquad = \ 512 - 1
\qquad \qquad \qquad \qquad \qquad = \ 511.
"This our answer."
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"Summarizing:"
\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ sum_{ n = 1 }^9 \ 2^{ n - 1 } \ = 511.
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"Note: There are other methods for doing this, using different"
"techniques, for example, one called changing the index of"
"summation. This is a good technique, and may be easier in"
"this case. The techniques displayed here illustrate alternatives"
"that can be powerful, and are quite frequently very convenient."
