Summing Up a Non-Geometric Series: Is It Possible?

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The discussion revolves around finding the sum of a non-geometric series represented by the formula ∑_{n=2}^∞ (50(-2)^{n-1}3^{n+2}/7^n). Participants suggest grouping terms to simplify the expression and explore the possibility of transforming it into a geometric series. One participant proposes a sum result of -8100/91, prompting questions about its correctness and the method of calculation. The conversation emphasizes the importance of recognizing patterns and using geometric series techniques to compute the sum. Ultimately, the discussion highlights the potential to apply learned geometric series methods to solve the problem.
RadiationX
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I need to find the sum of:

\sum_{n=2}^\infty\frac{50(-2)^{n-1}3^{n+2}}{7^n}


the only series that we've been taught to add up is geometric. the above series is not geometric,is it?
 
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RadiationX said:
I need to find the sum of:

\sum_{n=2}^\infty\frac{50(-2)^{n-1}3^{n+2}}{7^n}


the only series that we've been taught to add up is geometric. the above series is not geometric,is it?

Whatever it is, it has been disquised. Try gouping all the terms you can together to the power n, and reduce everything else and see what you get.
 
is \frac{-8100}{91} correct?
 
RadiationX said:
is \frac{-8100}{91} correct?

Looks OK to me
 
how would you compute this series by hand? i did this on a calculator. I don't see how i can combine the exponents given that they have different bases
 
RadiationX said:
how would you compute this series by hand? i did this on a calculator. I don't see how i can combine the exponents given that they have different bases
\sum_{n=2}^\infty\frac{50(-2)^{n-1}3^{n+2}}{7^n}

\frac{50 \bullet 9}{-2}\sum_{n=2}^\infty\frac{(-2)^{n}3^{n}}{7^n}

\frac{50 \bullet 9}{-2}\sum_{n=2}^\infty\left(\frac{-6}{7}\right)^n

\frac{50 \bullet 9}{-2}\left(\frac{-6}{7}\right)^2\sum_{n=0}^\infty\left(\frac{-6}{7}\right)^n

\frac{-50 \bullet 27}{49}\sum_{n=0}^\infty\left(\frac{-6}{7}\right)^n

If you have learned to do geometric sums, you can finish it. If you don't see how I changed the sum from n = 2 to n = 0, write out the first few terms of the sum to see how you can factor out the squared term.
 
beautiful. thank you!
 

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