# Homework Help: Some sums, don't sum up

1. Jul 20, 2006

### seaglespn

Some sums, don't sum up :)

I have a problem that require some math tricks, and after I tried to solve it myself I looked at the answer and I don't understand how this is done :
$$$\sum\limits_{k = 0}^n {\left( {\frac{2}{5}} \right)^k } + \sum\limits_{k = 0}^n {\left( {\frac{3}{5}} \right)^k } = \frac{5}{3}\left( {1 - \left( {\frac{2}{5}} \right)^{n + 1} } \right) + \frac{5}{2}\left( {1 - \left( {\frac{3}{5}} \right)^{n + 1} } \right)$$$

2. Jul 20, 2006

### nazzard

Hello seaglespn,

Hint: geometric series

$$\sum\limits_{k = 0}^n q^{k}=\frac{1-q^{n+1}}{1-q}$$

for $$|q|<1$$

Do you know how to prove this identity?

Although not necessary for solving this problem you might want to take a look at the infinite geometric series as well.

$$\sum\limits_{k = 0}^\infty q^{k}$$

for $$|q|<1$$

What would be the limit?

Regards,

nazzard

Last edited: Jul 20, 2006
3. Jul 20, 2006

### seaglespn

Ok, I have done the math, and I end up with the correct answer, after I wasn't so sure about the : $$$b_n = b_1 \frac{{q^n - 1}}{{q - 1}}$$$

Where the power of q must be the TOTAL number of elements...

Code (Text):

Sorry, my mistake... :smile:
The sum thends to a constant.... but that might be a definitions somewhere....
And it didn't rings any bell to me...
A constant "variable" due to q. :smile: .
Goofy me...

Thx for the help!

Regards,
seaglespn.

Last edited: Jul 20, 2006
4. Jul 20, 2006

### nazzard

$$\sum\limits_{k = 0}^\infty q^{k}=\lim_{\substack{n\rightarrow\infty}}\sum\limits_{k = 0}^{n} q^{k}=\lim_{\substack{n\rightarrow\infty}}\frac{1-q^{n+1}}{1-q}=?$$

Remember $$|q|<1$$

Last edited: Jul 20, 2006
5. Jul 20, 2006

### seaglespn

$$$\mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} = ?$$$

6. Jul 20, 2006

### seaglespn

$$$\mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}}$$$
?
Sorry about double post... my refresh is kinda slow

7. Jul 20, 2006

### nazzard

correct

Regards,

nazzard

8. Jul 20, 2006