How to solve this recurrence relation using generating functions: $a_n = 10 a_{n-1}-25 a_{n-2} +...












5












$begingroup$


How can we solve the following recurrence relation using GF?



$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$



I think that most of it is pretty straightforward. What really concerns me is this part



$5^n{n+2 choose 2}$



Problem further analyzed



After trying to create generating functions in the equation we end up in this



$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}} sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n$



Lets take this part:



$ sum_{n=2}^infty {n+2 choose 2}5^nx^n = sum_{n=2}^infty frac{(n+1)(n+2)}{2} 5^nx^n $



then?






sequences-and-series discrete-mathematics recurrence-relations generating-functions







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$endgroup$












  • $begingroup$
    $5^nbinom{n+2}2=5^ndfrac{(n+1)(n+2)}2$
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:13










  • $begingroup$
    It is multiplied. Edited the question so that it is clearer.
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:13
















5












$begingroup$


How can we solve the following recurrence relation using GF?



$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$



I think that most of it is pretty straightforward. What really concerns me is this part



$5^n{n+2 choose 2}$



Problem further analyzed



After trying to create generating functions in the equation we end up in this



$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}} sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n$



Lets take this part:



$ sum_{n=2}^infty {n+2 choose 2}5^nx^n = sum_{n=2}^infty frac{(n+1)(n+2)}{2} 5^nx^n $



then?






sequences-and-series discrete-mathematics recurrence-relations generating-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    $5^nbinom{n+2}2=5^ndfrac{(n+1)(n+2)}2$
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:13










  • $begingroup$
    It is multiplied. Edited the question so that it is clearer.
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:13














5












5








5


1



$begingroup$


How can we solve the following recurrence relation using GF?



$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$



I think that most of it is pretty straightforward. What really concerns me is this part



$5^n{n+2 choose 2}$



Problem further analyzed



After trying to create generating functions in the equation we end up in this



$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}} sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n$



Lets take this part:



$ sum_{n=2}^infty {n+2 choose 2}5^nx^n = sum_{n=2}^infty frac{(n+1)(n+2)}{2} 5^nx^n $



then?






sequences-and-series discrete-mathematics recurrence-relations generating-functions







share|cite|improve this question











$endgroup$




How can we solve the following recurrence relation using GF?



$a_n = 10 a_{n-1}-25 a_{n-2} + 5^n {n+2 choose 2}$ , for each $n>2, a_0 = 1, a_1 = 15$



I think that most of it is pretty straightforward. What really concerns me is this part



$5^n{n+2 choose 2}$



Problem further analyzed



After trying to create generating functions in the equation we end up in this



$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}} sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n$



Lets take this part:



$ sum_{n=2}^infty {n+2 choose 2}5^nx^n = sum_{n=2}^infty frac{(n+1)(n+2)}{2} 5^nx^n $



then?






sequences-and-series discrete-mathematics recurrence-relations generating-functions




sequences-and-series discrete-mathematics recurrence-relations generating-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 11 at 13:23







Dimitris Prasakis

















asked Feb 11 at 10:03









Dimitris PrasakisDimitris Prasakis

327




327












  • $begingroup$
    $5^nbinom{n+2}2=5^ndfrac{(n+1)(n+2)}2$
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:13










  • $begingroup$
    It is multiplied. Edited the question so that it is clearer.
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:13


















  • $begingroup$
    $5^nbinom{n+2}2=5^ndfrac{(n+1)(n+2)}2$
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:13










  • $begingroup$
    It is multiplied. Edited the question so that it is clearer.
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:13
















$begingroup$
$5^nbinom{n+2}2=5^ndfrac{(n+1)(n+2)}2$
$endgroup$
– Shubham Johri
Feb 11 at 10:13




$begingroup$
$5^nbinom{n+2}2=5^ndfrac{(n+1)(n+2)}2$
$endgroup$
– Shubham Johri
Feb 11 at 10:13












$begingroup$
It is multiplied. Edited the question so that it is clearer.
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:13




$begingroup$
It is multiplied. Edited the question so that it is clearer.
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:13










2 Answers
2






active

oldest

votes


















6












$begingroup$

Hint for finding the GF.



Note that $(n+1)(n+2)x^n=frac{d^2}{dx^2}left(x^{n+2}right)$
and therefore
$$sum_{n=2}^infty {n+2 choose 2}5^nx^n = frac{1}{2cdot 5^2}frac{d^2}{dx^2}left(
sum_{n=2}^infty (5x)^{n+2}
right).$$
Then recall that $sum_{n=0}^{infty}z^n=frac{1}{1-z}$.



Moreover in your attempt it should be
$$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}}x^n-25 sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
that is
$$sum_{n=2}^infty a_n x^n = 10x sum_{n=1}^infty {a_{n}}x^n-25x^2 sum_{n=0}^infty a_{n} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =sum_{n=0}^infty a_n x^n$?



Alternative way for finding $a_n$ without the GF.



The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation
$$z^2-10z+25=(z-5)^2=0$$
and non-homogeneous term $5^n {n+2 choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation.
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:38










  • $begingroup$
    @GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
    $endgroup$
    – Robert Z
    Feb 11 at 10:38










  • $begingroup$
    The indices can begin at $0.~ a_i=0$ for $i<0$.
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:42












  • $begingroup$
    @ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:45








  • 1




    $begingroup$
    @DimitrisPrasakis I edited my answer with an alternative method.
    $endgroup$
    – Robert Z
    Feb 11 at 11:02



















2












$begingroup$

$sum_{n=0}^infty {n+2 choose 2}5^nx^n = frac12sum_{n=0}^infty(n^2+3n+2)(5x)^n=frac12[sum_{n=0}^infty n^2y^n+3sum_{n=0}^infty ny^n+2sum_{n=0}^infty y^n]$



where $y=5x$. You will need the following series sums:



$(1)sum_{n=0}^infty n^2y^n=dfrac{y(1+y)}{(1-y)^3}\(2)sum_{n=0}^infty ny^n=dfrac y{(y-1)^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:39












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Hint for finding the GF.



Note that $(n+1)(n+2)x^n=frac{d^2}{dx^2}left(x^{n+2}right)$
and therefore
$$sum_{n=2}^infty {n+2 choose 2}5^nx^n = frac{1}{2cdot 5^2}frac{d^2}{dx^2}left(
sum_{n=2}^infty (5x)^{n+2}
right).$$
Then recall that $sum_{n=0}^{infty}z^n=frac{1}{1-z}$.



Moreover in your attempt it should be
$$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}}x^n-25 sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
that is
$$sum_{n=2}^infty a_n x^n = 10x sum_{n=1}^infty {a_{n}}x^n-25x^2 sum_{n=0}^infty a_{n} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =sum_{n=0}^infty a_n x^n$?



Alternative way for finding $a_n$ without the GF.



The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation
$$z^2-10z+25=(z-5)^2=0$$
and non-homogeneous term $5^n {n+2 choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation.
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:38










  • $begingroup$
    @GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
    $endgroup$
    – Robert Z
    Feb 11 at 10:38










  • $begingroup$
    The indices can begin at $0.~ a_i=0$ for $i<0$.
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:42












  • $begingroup$
    @ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:45








  • 1




    $begingroup$
    @DimitrisPrasakis I edited my answer with an alternative method.
    $endgroup$
    – Robert Z
    Feb 11 at 11:02
















6












$begingroup$

Hint for finding the GF.



Note that $(n+1)(n+2)x^n=frac{d^2}{dx^2}left(x^{n+2}right)$
and therefore
$$sum_{n=2}^infty {n+2 choose 2}5^nx^n = frac{1}{2cdot 5^2}frac{d^2}{dx^2}left(
sum_{n=2}^infty (5x)^{n+2}
right).$$
Then recall that $sum_{n=0}^{infty}z^n=frac{1}{1-z}$.



Moreover in your attempt it should be
$$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}}x^n-25 sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
that is
$$sum_{n=2}^infty a_n x^n = 10x sum_{n=1}^infty {a_{n}}x^n-25x^2 sum_{n=0}^infty a_{n} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =sum_{n=0}^infty a_n x^n$?



Alternative way for finding $a_n$ without the GF.



The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation
$$z^2-10z+25=(z-5)^2=0$$
and non-homogeneous term $5^n {n+2 choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation.
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:38










  • $begingroup$
    @GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
    $endgroup$
    – Robert Z
    Feb 11 at 10:38










  • $begingroup$
    The indices can begin at $0.~ a_i=0$ for $i<0$.
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:42












  • $begingroup$
    @ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:45








  • 1




    $begingroup$
    @DimitrisPrasakis I edited my answer with an alternative method.
    $endgroup$
    – Robert Z
    Feb 11 at 11:02














6












6








6





$begingroup$

Hint for finding the GF.



Note that $(n+1)(n+2)x^n=frac{d^2}{dx^2}left(x^{n+2}right)$
and therefore
$$sum_{n=2}^infty {n+2 choose 2}5^nx^n = frac{1}{2cdot 5^2}frac{d^2}{dx^2}left(
sum_{n=2}^infty (5x)^{n+2}
right).$$
Then recall that $sum_{n=0}^{infty}z^n=frac{1}{1-z}$.



Moreover in your attempt it should be
$$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}}x^n-25 sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
that is
$$sum_{n=2}^infty a_n x^n = 10x sum_{n=1}^infty {a_{n}}x^n-25x^2 sum_{n=0}^infty a_{n} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =sum_{n=0}^infty a_n x^n$?



Alternative way for finding $a_n$ without the GF.



The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation
$$z^2-10z+25=(z-5)^2=0$$
and non-homogeneous term $5^n {n+2 choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation.
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.






share|cite|improve this answer











$endgroup$



Hint for finding the GF.



Note that $(n+1)(n+2)x^n=frac{d^2}{dx^2}left(x^{n+2}right)$
and therefore
$$sum_{n=2}^infty {n+2 choose 2}5^nx^n = frac{1}{2cdot 5^2}frac{d^2}{dx^2}left(
sum_{n=2}^infty (5x)^{n+2}
right).$$
Then recall that $sum_{n=0}^{infty}z^n=frac{1}{1-z}$.



Moreover in your attempt it should be
$$sum_{n=2}^infty a_n x^n = 10 sum_{n=2}^infty {a_{n-1}}x^n-25 sum_{n=2}^infty a_{n-2} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
that is
$$sum_{n=2}^infty a_n x^n = 10x sum_{n=1}^infty {a_{n}}x^n-25x^2 sum_{n=0}^infty a_{n} x^n + sum_{n=2}^infty {n+2 choose 2}5^nx^n.$$
Can you take it from here and find $f(x) =sum_{n=0}^infty a_n x^n$?



Alternative way for finding $a_n$ without the GF.



The given recurrence is a non-homogeneous linear recurrence relation with constant coefficients with characteristic equation
$$z^2-10z+25=(z-5)^2=0$$
and non-homogeneous term $5^n {n+2 choose 2}$ which is a second degree polynomial multiplied by a power of $5$, which is the root of multiplicity $2$ of the characteristic equation.
Hence the general term of the recurrence with $a_0 = 1$, $a_1 = 15$ has the form
$$a_n= 5^n(An^4+Bn^3+Cn+D)$$
where $A,B,C,D$ are real constant to be determined.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Feb 11 at 11:02

























answered Feb 11 at 10:29









Robert ZRobert Z

102k1072145




102k1072145












  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:38










  • $begingroup$
    @GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
    $endgroup$
    – Robert Z
    Feb 11 at 10:38










  • $begingroup$
    The indices can begin at $0.~ a_i=0$ for $i<0$.
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:42












  • $begingroup$
    @ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:45








  • 1




    $begingroup$
    @DimitrisPrasakis I edited my answer with an alternative method.
    $endgroup$
    – Robert Z
    Feb 11 at 11:02


















  • $begingroup$
    Great answer, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:38










  • $begingroup$
    @GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
    $endgroup$
    – Robert Z
    Feb 11 at 10:38










  • $begingroup$
    The indices can begin at $0.~ a_i=0$ for $i<0$.
    $endgroup$
    – Shubham Johri
    Feb 11 at 10:42












  • $begingroup$
    @ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:45








  • 1




    $begingroup$
    @DimitrisPrasakis I edited my answer with an alternative method.
    $endgroup$
    – Robert Z
    Feb 11 at 11:02
















$begingroup$
Great answer, thank you!
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:38




$begingroup$
Great answer, thank you!
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:38












$begingroup$
@GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
$endgroup$
– Robert Z
Feb 11 at 10:38




$begingroup$
@GEdgar I would say an algebraic equation for $f(x) =sum_{n=0}^infty a_n x^n$.
$endgroup$
– Robert Z
Feb 11 at 10:38












$begingroup$
The indices can begin at $0.~ a_i=0$ for $i<0$.
$endgroup$
– Shubham Johri
Feb 11 at 10:42






$begingroup$
The indices can begin at $0.~ a_i=0$ for $i<0$.
$endgroup$
– Shubham Johri
Feb 11 at 10:42














$begingroup$
@ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:45






$begingroup$
@ShubhamJohri the description of the problem says foreach $n geq 2$ given that $a_0 = 1, a_1 = 15$. Thus the indices start at 2 and then we got to degrade them to $0$, so that we use the definition of generating functions (where indices start from $0$)
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:45






1




1




$begingroup$
@DimitrisPrasakis I edited my answer with an alternative method.
$endgroup$
– Robert Z
Feb 11 at 11:02




$begingroup$
@DimitrisPrasakis I edited my answer with an alternative method.
$endgroup$
– Robert Z
Feb 11 at 11:02











2












$begingroup$

$sum_{n=0}^infty {n+2 choose 2}5^nx^n = frac12sum_{n=0}^infty(n^2+3n+2)(5x)^n=frac12[sum_{n=0}^infty n^2y^n+3sum_{n=0}^infty ny^n+2sum_{n=0}^infty y^n]$



where $y=5x$. You will need the following series sums:



$(1)sum_{n=0}^infty n^2y^n=dfrac{y(1+y)}{(1-y)^3}\(2)sum_{n=0}^infty ny^n=dfrac y{(y-1)^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:39
















2












$begingroup$

$sum_{n=0}^infty {n+2 choose 2}5^nx^n = frac12sum_{n=0}^infty(n^2+3n+2)(5x)^n=frac12[sum_{n=0}^infty n^2y^n+3sum_{n=0}^infty ny^n+2sum_{n=0}^infty y^n]$



where $y=5x$. You will need the following series sums:



$(1)sum_{n=0}^infty n^2y^n=dfrac{y(1+y)}{(1-y)^3}\(2)sum_{n=0}^infty ny^n=dfrac y{(y-1)^2}$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:39














2












2








2





$begingroup$

$sum_{n=0}^infty {n+2 choose 2}5^nx^n = frac12sum_{n=0}^infty(n^2+3n+2)(5x)^n=frac12[sum_{n=0}^infty n^2y^n+3sum_{n=0}^infty ny^n+2sum_{n=0}^infty y^n]$



where $y=5x$. You will need the following series sums:



$(1)sum_{n=0}^infty n^2y^n=dfrac{y(1+y)}{(1-y)^3}\(2)sum_{n=0}^infty ny^n=dfrac y{(y-1)^2}$






share|cite|improve this answer









$endgroup$



$sum_{n=0}^infty {n+2 choose 2}5^nx^n = frac12sum_{n=0}^infty(n^2+3n+2)(5x)^n=frac12[sum_{n=0}^infty n^2y^n+3sum_{n=0}^infty ny^n+2sum_{n=0}^infty y^n]$



where $y=5x$. You will need the following series sums:



$(1)sum_{n=0}^infty n^2y^n=dfrac{y(1+y)}{(1-y)^3}\(2)sum_{n=0}^infty ny^n=dfrac y{(y-1)^2}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Feb 11 at 10:33









Shubham JohriShubham Johri

5,613818




5,613818












  • $begingroup$
    This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:39


















  • $begingroup$
    This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
    $endgroup$
    – Dimitris Prasakis
    Feb 11 at 10:39
















$begingroup$
This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:39




$begingroup$
This is a working approach. @Robert Z provided a faster one though. In any case, thank you!
$endgroup$
– Dimitris Prasakis
Feb 11 at 10:39


















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