Mixing
Manipulation of Power Series [duplicate]
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Proof of the equality $sumlimits_k=1^infty frack^22^k = 6$
8 answers
Show that ... $$sum_n=1^infty fracn^2 2^n = 6$$
How would one go about in showing that the above power series equals to 6? I would assume that it has something to do with the following properties such as ...
f(x) = $sum Cn·X^n$, g(x) = $sum Dn·X^n$
f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$
f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$
$sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$
if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$
if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$
$sum_n=b^infty Cn·X^n$ is continuous on Interval of Convergence.
This is the completed work so far ...
$$sum_n=1^infty fracn^2 2^n = sum_n=1^infty (1/2)^n ·n^2$$
let f(x) = $$ sum_n=1^infty X^nn^2$$
then $$ int f(x) = sum_n=1^infty frac X^n+1 n+1 · n^2$$
I assume the next step should be to cancel $n^2$ and n+1, but I'm not quite getting the connection. How would you proceed with this?
calculus sequences-and-series power-series
asked Jul 30 at 5:34
LeviRivai
61
marked as duplicate by Jyrki Lahtonen, Community♦ Jul 30 at 5:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
1
down vote
favorite
This question already has an answer here:
Proof of the equality $sumlimits_k=1^infty frack^22^k = 6$
8 answers
Show that ... $$sum_n=1^infty fracn^2 2^n = 6$$
How would one go about in showing that the above power series equals to 6? I would assume that it has something to do with the following properties such as ...
f(x) = $sum Cn·X^n$, g(x) = $sum Dn·X^n$
f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$
f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$
$sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$
if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$
if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$
$sum_n=b^infty Cn·X^n$ is continuous on Interval of Convergence.
This is the completed work so far ...
$$sum_n=1^infty fracn^2 2^n = sum_n=1^infty (1/2)^n ·n^2$$
let f(x) = $$ sum_n=1^infty X^nn^2$$
then $$ int f(x) = sum_n=1^infty frac X^n+1 n+1 · n^2$$
I assume the next step should be to cancel $n^2$ and n+1, but I'm not quite getting the connection. How would you proceed with this?
calculus sequences-and-series power-series
asked Jul 30 at 5:34
LeviRivai
61
marked as duplicate by Jyrki Lahtonen, Community♦ Jul 30 at 5:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Huh? $$
– copper.hat
Jul 30 at 5:38
Approach0 finds a few copies and several closely related sums.
– Jyrki Lahtonen
Jul 30 at 5:53
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Proof of the equality $sumlimits_k=1^infty frack^22^k = 6$
8 answers
Show that ... $$sum_n=1^infty fracn^2 2^n = 6$$
How would one go about in showing that the above power series equals to 6? I would assume that it has something to do with the following properties such as ...
f(x) = $sum Cn·X^n$, g(x) = $sum Dn·X^n$
f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$
f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$
$sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$
if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$
if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$
$sum_n=b^infty Cn·X^n$ is continuous on Interval of Convergence.
This is the completed work so far ...
$$sum_n=1^infty fracn^2 2^n = sum_n=1^infty (1/2)^n ·n^2$$
let f(x) = $$ sum_n=1^infty X^nn^2$$
then $$ int f(x) = sum_n=1^infty frac X^n+1 n+1 · n^2$$
I assume the next step should be to cancel $n^2$ and n+1, but I'm not quite getting the connection. How would you proceed with this?
calculus sequences-and-series power-series
asked Jul 30 at 5:34
LeviRivai
61
This question already has an answer here:
Proof of the equality $sumlimits_k=1^infty frack^22^k = 6$
8 answers
Show that ... $$sum_n=1^infty fracn^2 2^n = 6$$
How would one go about in showing that the above power series equals to 6? I would assume that it has something to do with the following properties such as ...
f(x) = $sum Cn·X^n$, g(x) = $sum Dn·X^n$
f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$
f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$
$sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$
if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$
if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$
$sum_n=b^infty Cn·X^n$ is continuous on Interval of Convergence.
This is the completed work so far ...
$$sum_n=1^infty fracn^2 2^n = sum_n=1^infty (1/2)^n ·n^2$$
let f(x) = $$ sum_n=1^infty X^nn^2$$
then $$ int f(x) = sum_n=1^infty frac X^n+1 n+1 · n^2$$
I assume the next step should be to cancel $n^2$ and n+1, but I'm not quite getting the connection. How would you proceed with this?
This question already has an answer here:
Proof of the equality $sumlimits_k=1^infty frack^22^k = 6$
8 answers
calculus sequences-and-series power-series
asked Jul 30 at 5:34
LeviRivai
61
asked Jul 30 at 5:34
LeviRivai
61
asked Jul 30 at 5:34
LeviRivai
61
asked Jul 30 at 5:34
LeviRivai
61
61
marked as duplicate by Jyrki Lahtonen, Community♦ Jul 30 at 5:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Jyrki Lahtonen, Community♦ Jul 30 at 5:59
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
Huh? $$
– copper.hat
Jul 30 at 5:38
Approach0 finds a few copies and several closely related sums.
– Jyrki Lahtonen
Jul 30 at 5:53
add a comment |Â
Huh? $$
– copper.hat
Jul 30 at 5:38
Approach0 finds a few copies and several closely related sums.
– Jyrki Lahtonen
Jul 30 at 5:53
Huh? $$
– copper.hat
Jul 30 at 5:38
Huh? $$
– copper.hat
Jul 30 at 5:38
Approach0 finds a few copies and several closely related sums.
– Jyrki Lahtonen
Jul 30 at 5:53
Approach0 finds a few copies and several closely related sums.
– Jyrki Lahtonen
Jul 30 at 5:53
add a comment |Â
2 Answers
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Hinte: Try starting with $sum_n=0^infty x^n = 1 over 1-x$ for $|x|<1$ and differentiate both sides. Multiply the result by $x$ and repeat. Adjust as necessary. Rinse & dry.
answered Jul 30 at 5:42
copper.hat
122k557156
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Consider$$S_k=sum_n=1^k n^2 x^n=sum_n=1^k [n(n-1)+n] x^n=sum_n=1^k n(n-1) x^n+sum_n=1^k n x^n$$ that is to say
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hinte: Try starting with $sum_n=0^infty x^n = 1 over 1-x$ for $|x|<1$ and differentiate both sides. Multiply the result by $x$ and repeat. Adjust as necessary. Rinse & dry.
answered Jul 30 at 5:42
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
Hinte: Try starting with $sum_n=0^infty x^n = 1 over 1-x$ for $|x|<1$ and differentiate both sides. Multiply the result by $x$ and repeat. Adjust as necessary. Rinse & dry.
answered Jul 30 at 5:42
copper.hat
122k557156
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hinte: Try starting with $sum_n=0^infty x^n = 1 over 1-x$ for $|x|<1$ and differentiate both sides. Multiply the result by $x$ and repeat. Adjust as necessary. Rinse & dry.
answered Jul 30 at 5:42
copper.hat
122k557156
Hinte: Try starting with $sum_n=0^infty x^n = 1 over 1-x$ for $|x|<1$ and differentiate both sides. Multiply the result by $x$ and repeat. Adjust as necessary. Rinse & dry.
answered Jul 30 at 5:42
copper.hat
122k557156
answered Jul 30 at 5:42
copper.hat
122k557156
answered Jul 30 at 5:42
copper.hat
122k557156
answered Jul 30 at 5:42
copper.hat
122k557156
122k557156
add a comment |Â
add a comment |Â
up vote
0
down vote
Consider$$S_k=sum_n=1^k n^2 x^n=sum_n=1^k [n(n-1)+n] x^n=sum_n=1^k n(n-1) x^n+sum_n=1^k n x^n$$ that is to say
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
Consider$$S_k=sum_n=1^k n^2 x^n=sum_n=1^k [n(n-1)+n] x^n=sum_n=1^k n(n-1) x^n+sum_n=1^k n x^n$$ that is to say
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Consider$$S_k=sum_n=1^k n^2 x^n=sum_n=1^k [n(n-1)+n] x^n=sum_n=1^k n(n-1) x^n+sum_n=1^k n x^n$$ that is to say
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
Consider$$S_k=sum_n=1^k n^2 x^n=sum_n=1^k [n(n-1)+n] x^n=sum_n=1^k n(n-1) x^n+sum_n=1^k n x^n$$ that is to say
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
answered Jul 30 at 5:57
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
Huh? $$
– copper.hat
Jul 30 at 5:38
Approach0 finds a few copies and several closely related sums.
– Jyrki Lahtonen
Jul 30 at 5:53