Sum of a non-geometric series with exponential terms

[RadiationX]

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?

 

[OlderDan]

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.

 

[RadiationX]

is $$\frac{-8100}{91}$$ correct?

 

[OlderDan]

is $$\frac{-8100}{91}$$ correct?

Looks OK to me

 

[RadiationX]

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

 

[OlderDan]

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.

 

[RadiationX]

beautiful. thank you!