# Thread: Problem on function definition

1. ## Problem on function definition

Given the function is defined by, f(n)=1+0.62+0.63+0.64+0.65+...+0.6n-1+0.6n
How can I define it in a more definite way and shorter? Help please!

2. ## Re: Problem on function definition

Originally Posted by yugimutoshung
Given the function is defined by, f(n)=1+0.62+0.63+0.64+0.65+...+0.6n-1+0.6n
How can I define it in a more definite way and shorter? Help please!

$\displaystyle f(n)=1 + r + {r^2} + \cdots + {r^n} = \frac{{1 - {r^{n + 1}}}}{{1 - r}}$

3. ## Re: Problem on function definition

You could write it as \displaystyle \displaystyle \begin{align*} f(n) = \sum_{k = 0}^{n} \left( \frac{3}{5} \right) ^k - \frac{3}{5} \end{align*}, and this sum is a finite geometric series, and so its sum can be found using the finite geometric series formula.

4. ## Re: Problem on function definition

OMG thank you so much! I tried numerical method but failed for hours

5. ## Re: Problem on function definition

what about f(n) = 1 + 2(0.6)^1 + 2(0.6)^2 + (0.6)^3 + ... + 2(0.6)^n ?

6. ## Re: Problem on function definition

Originally Posted by yugimutoshung
what about f(n) = 1 + 2(0.6)^1 + 2(0.6)^2 + (0.6)^3 + ... + 2(0.6)^n ?
What do you have to do in order to turn it into a geometric series?

7. ## Re: Problem on function definition

I forgot the formula... So how can I do it?

8. ## Re: Problem on function definition

Originally Posted by yugimutoshung
what about f(n) = 1 + 2(0.6)^1 + 2(0.6)^2 + (0.6)^3 + ... + 2(0.6)^n ?
The regularity denoted by ... is not obvious. In other words, it is not clear what the skipped terms are. You have factor 2 in terms where 0.6 is raised to powers 1, 2 and n, but not 0 or 3.