Permutations and Combinations

Question

Find the sum to indicated number of terms in each of the geometric progressions:

$open parentheses 1 fifth plus 2 over 5 squared plus 3 over 5 cubed close parentheses comma space space open parentheses 1 over 5 to the power of 4 plus 2 over 5 to the power of 5 plus 3 over 5 to the power of 6 close parentheses comma space open parentheses 1 over 5 to the power of 7 plus 2 over 5 to the power of 8 plus 3 over 5 to the power of 9 close parentheses comma space...... semicolon 7$

Solution

The given G.P. is:
 $space space open parentheses 1 fifth plus 2 over 5 squared plus 3 over 5 cubed close parentheses comma space open parentheses 1 over 5 to the power of 4 plus 2 over 5 to the power of 5 plus 3 over 5 to the power of 6 close parentheses comma space open parentheses 1 over 5 to the power of 7 plus 2 over 5 to the power of 8 plus 3 over 5 to the power of 9 close parentheses comma space..................$
or $open parentheses 1 fifth plus 2 over 5 squared plus 3 over 5 cubed close parentheses comma space space 1 over 5 cubed open parentheses 1 fifth plus 2 over 5 squared plus 3 over 5 cubed close parentheses comma space 1 over 5 to the power of 6 open parentheses 1 fifth plus 2 over 5 squared plus 3 over 5 cubed close parentheses comma space......$
Here, $straight a space equals space 1 fifth plus 2 over 5 squared plus 3 over 5 cubed$ and $straight r space equals space fraction numerator begin display style 1 over 5 cubed end style open parentheses begin display style 1 fifth end style plus begin display style 2 over 5 squared end style plus begin display style 3 over 5 cubed end style close parentheses over denominator open parentheses begin display style 1 fifth end style plus begin display style 2 over 5 squared end style plus begin display style 3 over 5 cubed end style close parentheses end fraction space equals space 1 over 5 cubed$
$straight S subscript 7 space equals space fraction numerator straight a left parenthesis 1 minus straight r to the power of 7 right parenthesis over denominator 1 minus straight r end fraction$ (∵ r < 1)
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