How do I solve this tricky math problem involving sums?

[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 :
<br /> \[<br /> \sum\limits_{k = 0}^n {\left( {\frac{2}{5}} \right)^k } + \sum\limits_{k = 0}^n {\left( {\frac{3}{5}} \right)^k } = \frac{5}{3}\left( {1 - \left( {\frac{2}{5}} \right)^{n + 1} } \right) + \frac{5}{2}\left( {1 - \left( {\frac{3}{5}} \right)^{n + 1} } \right)<br /> \]<br />

An advice pls, thx!

 

[nazzard]

Hello seaglespn,

Hint: geometric series

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

for |q|&lt;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|&lt;1

What would be the limit?

Regards,

nazzard

 

[seaglespn]

Hello seaglespn,

Hint: geometric series

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

for |q|&lt;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|&lt;1

What would be the limit?

Regards,

nazzard

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

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

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.

 

[nazzard]

And about the sum which tends to infinit the limit would be 0 if |x|<1 , else it would be infinite...

Try again please

\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|&lt;1

 

[seaglespn]

<br /> \[<br /> \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} = ?<br /> \]<br />

 

[seaglespn]

<br /> \[<br /> \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} \]<br /> <br />
?
Sorry about double post... my refresh is kinda slow

 

[nazzard]

<br /> \[<br /> \mathop {\lim }\limits_{n \to \infty } \frac{{1 - q^{n + 1} }}{{1 - q}} = \frac{1}{{1 - q}} \]<br /> <br />
?

correct

Regards,

nazzard

 

[seaglespn]

Thanks for your help @nazzard...

Cheers!