Find constant k with the probability mass function of a geometric series












0












$begingroup$


Ok so I'm having some trouble getting this done. I think I've got the answer but I'm not really sure on how I could verify that the result I got is actually the correct answer.



So here's what the exercise says:



$f(x) = k(frac{1}{5})^{2x+3}$



where a discrete random variable $X = 0,1,2,3,....$



So I know this is a function for a geometric series given that the range for X is infinite.



Here's what I did:




  1. $sumlimits_{i = 0}^infty k(frac{1}{5})^{2x + 3} = 1$

  2. $frac{k}{125}sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = 1$

  3. $frac{k}{125}(frac{1}{1-frac{1}{5}}) = 1$

  4. $frac{k}{125}(frac{5}{4}) = 1$

  5. $frac{k}{100} = 1$

  6. $k = 100$


So how do I know that is ok, I thought about using limits but I'm not sure if that is correct.






probability probability-distributions







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$endgroup$












  • $begingroup$
    $sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = sumlimits_{i = 0}^infty (frac{1}{25})^{x} = frac{25}{24}$
    $endgroup$
    – Henry
    Sep 30 '15 at 22:29


















0












$begingroup$


Ok so I'm having some trouble getting this done. I think I've got the answer but I'm not really sure on how I could verify that the result I got is actually the correct answer.



So here's what the exercise says:



$f(x) = k(frac{1}{5})^{2x+3}$



where a discrete random variable $X = 0,1,2,3,....$



So I know this is a function for a geometric series given that the range for X is infinite.



Here's what I did:




  1. $sumlimits_{i = 0}^infty k(frac{1}{5})^{2x + 3} = 1$

  2. $frac{k}{125}sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = 1$

  3. $frac{k}{125}(frac{1}{1-frac{1}{5}}) = 1$

  4. $frac{k}{125}(frac{5}{4}) = 1$

  5. $frac{k}{100} = 1$

  6. $k = 100$


So how do I know that is ok, I thought about using limits but I'm not sure if that is correct.






probability probability-distributions







share|cite|improve this question









$endgroup$












  • $begingroup$
    $sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = sumlimits_{i = 0}^infty (frac{1}{25})^{x} = frac{25}{24}$
    $endgroup$
    – Henry
    Sep 30 '15 at 22:29
















0












0








0





$begingroup$


Ok so I'm having some trouble getting this done. I think I've got the answer but I'm not really sure on how I could verify that the result I got is actually the correct answer.



So here's what the exercise says:



$f(x) = k(frac{1}{5})^{2x+3}$



where a discrete random variable $X = 0,1,2,3,....$



So I know this is a function for a geometric series given that the range for X is infinite.



Here's what I did:




  1. $sumlimits_{i = 0}^infty k(frac{1}{5})^{2x + 3} = 1$

  2. $frac{k}{125}sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = 1$

  3. $frac{k}{125}(frac{1}{1-frac{1}{5}}) = 1$

  4. $frac{k}{125}(frac{5}{4}) = 1$

  5. $frac{k}{100} = 1$

  6. $k = 100$


So how do I know that is ok, I thought about using limits but I'm not sure if that is correct.






probability probability-distributions







share|cite|improve this question









$endgroup$




Ok so I'm having some trouble getting this done. I think I've got the answer but I'm not really sure on how I could verify that the result I got is actually the correct answer.



So here's what the exercise says:



$f(x) = k(frac{1}{5})^{2x+3}$



where a discrete random variable $X = 0,1,2,3,....$



So I know this is a function for a geometric series given that the range for X is infinite.



Here's what I did:




  1. $sumlimits_{i = 0}^infty k(frac{1}{5})^{2x + 3} = 1$

  2. $frac{k}{125}sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = 1$

  3. $frac{k}{125}(frac{1}{1-frac{1}{5}}) = 1$

  4. $frac{k}{125}(frac{5}{4}) = 1$

  5. $frac{k}{100} = 1$

  6. $k = 100$


So how do I know that is ok, I thought about using limits but I'm not sure if that is correct.






probability probability-distributions




probability probability-distributions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 30 '15 at 22:22









ArgusArgus

1947




1947












  • $begingroup$
    $sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = sumlimits_{i = 0}^infty (frac{1}{25})^{x} = frac{25}{24}$
    $endgroup$
    – Henry
    Sep 30 '15 at 22:29




















  • $begingroup$
    $sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = sumlimits_{i = 0}^infty (frac{1}{25})^{x} = frac{25}{24}$
    $endgroup$
    – Henry
    Sep 30 '15 at 22:29


















$begingroup$
$sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = sumlimits_{i = 0}^infty (frac{1}{25})^{x} = frac{25}{24}$
$endgroup$
– Henry
Sep 30 '15 at 22:29






$begingroup$
$sumlimits_{i = 0}^infty (frac{1}{5})^{2x} = sumlimits_{i = 0}^infty (frac{1}{25})^{x} = frac{25}{24}$
$endgroup$
– Henry
Sep 30 '15 at 22:29












1 Answer
1






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oldest

votes


















0












$begingroup$

Observe
begin{align}
sum_{X=0}^{infty}kleft(frac{1}{5}right)^{2X+3}&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{5}right)^{2X}\
&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{25}right)^X\
&=frac{k}{125}cdotfrac{1}{1-frac{1}{25}}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Crap, I forgot about that small detail.
    $endgroup$
    – Argus
    Sep 30 '15 at 22:30











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Observe
begin{align}
sum_{X=0}^{infty}kleft(frac{1}{5}right)^{2X+3}&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{5}right)^{2X}\
&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{25}right)^X\
&=frac{k}{125}cdotfrac{1}{1-frac{1}{25}}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Crap, I forgot about that small detail.
    $endgroup$
    – Argus
    Sep 30 '15 at 22:30
















0












$begingroup$

Observe
begin{align}
sum_{X=0}^{infty}kleft(frac{1}{5}right)^{2X+3}&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{5}right)^{2X}\
&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{25}right)^X\
&=frac{k}{125}cdotfrac{1}{1-frac{1}{25}}
end{align}






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Crap, I forgot about that small detail.
    $endgroup$
    – Argus
    Sep 30 '15 at 22:30














0












0








0





$begingroup$

Observe
begin{align}
sum_{X=0}^{infty}kleft(frac{1}{5}right)^{2X+3}&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{5}right)^{2X}\
&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{25}right)^X\
&=frac{k}{125}cdotfrac{1}{1-frac{1}{25}}
end{align}






share|cite|improve this answer









$endgroup$



Observe
begin{align}
sum_{X=0}^{infty}kleft(frac{1}{5}right)^{2X+3}&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{5}right)^{2X}\
&=frac{k}{125}sum_{X=0}^{infty}left(frac{1}{25}right)^X\
&=frac{k}{125}cdotfrac{1}{1-frac{1}{25}}
end{align}







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 30 '15 at 22:27









Ángel Mario GallegosÁngel Mario Gallegos

18.5k11230




18.5k11230












  • $begingroup$
    Crap, I forgot about that small detail.
    $endgroup$
    – Argus
    Sep 30 '15 at 22:30


















  • $begingroup$
    Crap, I forgot about that small detail.
    $endgroup$
    – Argus
    Sep 30 '15 at 22:30
















$begingroup$
Crap, I forgot about that small detail.
$endgroup$
– Argus
Sep 30 '15 at 22:30




$begingroup$
Crap, I forgot about that small detail.
$endgroup$
– Argus
Sep 30 '15 at 22:30


















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