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 November 11th, 2013, 08:45 PM #1 Senior Member   Joined: Jan 2009 Posts: 345 Thanks: 3 Sum of this expression What's the summation of the following expression; $\text{\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^{k} \left(\frac{1}{4}\right)^{n-k}}$ The solution is said to $\text{2\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)}$ But I'm getting $\text{\left(\frac{1}{4} \right)^{n}\left(2^{n+3}-1\right)}$. How is this possible?
 November 12th, 2013, 08:04 AM #2 Senior Member   Joined: Feb 2013 Posts: 281 Thanks: 0 Re: Sum of this expression My guesswork is that you didn't apply the geometrical progression formula corerctly. Did you consider that k starts with 1 instead of 0?
 November 12th, 2013, 12:44 PM #3 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Sum of this expression $Sum= \dfrac{a(r^m - 1)}{r-1}$, where $a$ is the first term of the geometric series, $m$ is the number of terms of the series and $r$ is the common ratio. For your problem, $a= 2(\frac{1}{4})^n , m = n+3 , r = 2$. Plug these in the formula to convince yourself of the final answer.
November 12th, 2013, 07:39 PM   #4
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Re: Sum of this expression

Quote:
 Originally Posted by AfroMike $Sum= \dfrac{a(r^m - 1)}{r-1}$, where $a$ is the first term of the geometric series, $m$ is the number of terms of the series and $r$ is the common ratio. For your problem, $a= 2(\frac{1}{4})^n , m = n+3 , r = 2$. Plug these in the formula to convince yourself of the final answer.
I don't know where you got the 2 from but here's what I'm doing. Maybe you can catch my mistake;

$\text{\sum_{k=1}^{n+3}\left(\frac{1}{2}\right)^k \left(\frac{1}{4}\right)^n \left(\frac{1}{4}\right)^{-k} \rightarrow \left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3}\left(\frac{1}{4}\right)^k 2^k \left(\frac{1}{4}\right)^{-k} \rightarrow \left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3} 2^k \rightarrow \left(\frac{1}{4}\right)^n \left(\frac{1-2^{n+3}}{1-2}\right) \rightarrow \left(\frac{1}{4}\right)^n \left(2^{n+3}-1\right)}$

 November 14th, 2013, 05:54 AM #5 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Sum of this expression Your mistake is that $\sum_{k=1}^{n+3}2^k = \dfrac{2(2^{n+3} - 1)}{2 - 1}$
November 17th, 2013, 01:38 AM   #6
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Re: Sum of this expression

Quote:
 Originally Posted by AfroMike Your mistake is that $\sum_{k=1}^{n+3}2^k = \dfrac{2(2^{n+3} - 1)}{2 - 1}$
That doesn't even look like the expression I wrote down

 November 17th, 2013, 08:39 AM #7 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 Re: Sum of this expression Sivela, you mean to say that within the past week you haven't even detected your error?
November 18th, 2013, 06:55 PM   #8
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Re: Sum of this expression

Quote:
 Originally Posted by AfroMike Sivela, you mean to say that within the past week you haven't even detected your error?
That's beside the point. Your expression looks like nothing I wrote down.
My mistake was forgetting to multiply my result by 2, the first term as the formula shows. I don't know how you could have missed this.

November 19th, 2013, 06:07 AM   #9
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Re: Sum of this expression

Ok, Sivela. It appeared you were being adamant. I'll quote here where I detected your error
Quote:
 Originally Posted by sivela $\left(\frac{1}{4}\right)^n \sum_{k=1}^{n+3} 2^k \rightarrow \left(\frac{1}{4}\right)^n \left(\frac{1-2^{n+3}}{1-2}\right)$
You must not take my remark to heart - rather be more observant. Cheers.

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