Question

Write the sum using summation notation, assuming the suggested pattern continues.

Answer 1

the first six terms of the sequence $\bf a_1=-1\qquad a_n=2\times a_{n-1}$

$\bf \begin{array}{ccll}n&term&value\\\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\1&a_1&-1\\2&a_2&2\times a_{2-1}\\&&2\times a_1\\&&-2\\3&a_3&2\times a_{3-1}\\&&2\times a_2\\&&-4\\4&a_4&2\times a_{4-1}\\&&2\times a_3\\&&-8\\5&a_5&2\times a_{5-1}\\&&2\times a_4\\&&-16\\6&a_6&2\times a_{-1}\\&&2\times a_5\\&&-32\end{array}$

now on this new edited version

$\bf -8~~,~~\stackrel{-8+5}{-3}~~,~~\stackrel{-3+5}{2}~~,~~\stackrel{2+5}{7}....67$

so, as you can see, the "common difference" of this arithmetic sequence is d = 5, hmmm what term is 67 anyway?

$\bf n^{th}\textit{ term of an arithmetic sequence}\\\\a_n=a_1+(n-1)d\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\d=\textit{common difference}\\a_1=-8\\d=5\\a_n=67\end{cases}\\\\\\67=-8+(n-1)5\implies 67=-8-5+5n\implies 67=-13+5n\\\\\\80=5n\implies \cfrac{80}{5}=n\implies 16=n\qquad ~~~~~~~~~~\boxed{\sum\limits_{i=0}^{15}~-8+5i}$