Question

How do I find the closed form for the nth partial sum of this series: 2/(5^(k-1))?

Answer 1

lookup geometric series partial sum formula

Answer 2

Hi there!

See for solving the partial sum... compare the general term with geometric series as @rational suggested

Answer 3

Remember that partial sum \(S_n\) can be calculated using :

\[\large{S_n = \Sum_{k=1}^{n} \cfrac{2}{5^{k-1}}}\]

Answer 4

\[\large{S_n = \sum_{k=1}^{n} \cfrac{2}{5^{k-1}}}\]

Answer 5

Can you give it a try @Dragondd320 ?

Answer 6

The answer I am given is \[(5/2)(1-(1/5)^n)\]

Answer 7

And I cannot seem to replicate that answer, I attribute it mainly to lack of sleep but I just need a push in the right direction and I can go from there.

Answer 8

See, the general term of a geometric series is:

\[\large{ar^{k-1}}\]

The sum of this geometric series is given by:

\[\large{S_n = \sum_{k = 1}^{n} (ar^{k-1}) = \cfrac{a(1-r^{n})}{1 - r}}\]

Answer 9

Now, try comparing the general term \(ar^{k-1}\) with \(\cfrac{2}{5^{k-1}}\)

Answer 10

I'm currently stuck at:

Is the next step:

Logs always mess me up. So that's why I am stuck.