# Calculate value / evaluate Series @ infinity

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:

$$\sum_{n = 2}^{\infty} 7 * \frac{-5^{n}}{2^{3n-2}}$$

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 Do you want $2^{3n-2}$ in the denominator? I imagine this is what was intended. – David Mitra Feb 15 '12 at 13:12 Thanks its fixed now. – kellax Feb 15 '12 at 13:18 Can you write this as a "geometric series"? Is "geometric series" a topic in your course? – GEdgar Feb 15 '12 at 13:33

Assuming you have $(-5)^n$ upstairs in the sum.
$$7\cdot{ (-5)^n\over 2^{3n-2}} = 7\cdot{ (-5)^n\over2^{-2} 2^{3n } } = {4\cdot 7}\cdot{ (-5)^n\over ({2^3})^{ n }} ={28}\cdot{ \Bigl({-5\over 8}\Bigr)^{ n }}.$$
Then $$\sum_{n=2}^\infty 7\cdot{ (-5)^n\over 2^{3n-2} } = \sum_{n=2}^\infty {28}\cdot{ \Bigl({-5\over 8}\Bigr)^{ n }} ={28}\cdot {(-5/8)^2\over 1-(-5/8)} ={28}\cdot{25/64\over 13/8 } ={28}\cdot{25\over 13\cdot8 } ={{25\cdot 7\over 13\cdot 2}}.$$