I got this on my quiz yesterday it was the only thing that i could not solve.
We were asked to evaluate the following series:
$ \sum_{n = 2}^{\infty} 7 * \frac{-5^{n}}{2^{3n-2}} $
I got this on my quiz yesterday it was the only thing that i could not solve.
We were asked to evaluate the following series:
$ \sum_{n = 2}^{\infty} 7 * \frac{-5^{n}}{2^{3n-2}} $
Assuming you have $(-5)^n$ upstairs in the sum.
Try to write the terms of this Geometric series in standard form:
$ 7\cdot{ (-5)^n\over 2^{3n-2}} = 7\cdot{ (-5)^n\over2^{-2} 2^{3n } } = {4\cdot 7}\cdot{ (-5)^n\over ({2^3})^{ n }} ={28}\cdot{ \Bigl({-5\over 8}\Bigr)^{ n }}. $
Then $ \sum_{n=2}^\infty 7\cdot{ (-5)^n\over 2^{3n-2} } = \sum_{n=2}^\infty {28}\cdot{ \Bigl({-5\over 8}\Bigr)^{ n }} ={28}\cdot {(-5/8)^2\over 1-(-5/8)} ={28}\cdot{25/64\over 13/8 } ={28}\cdot{25\over 13\cdot8 } ={{25\cdot 7\over 13\cdot 2}}. $
(The sum of a convergent Geometric series is the first term divided by (1- ratio)).