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) = $sum Dn·X^n$

1. f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$




2. f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$




3. $sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$




4. if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$




5. if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$




6. $sum_n=b^infty Cn·X^n$ 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

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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 | 

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) = $sum Dn·X^n$

1. f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$




2. f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$




3. $sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$




4. if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$




5. if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$




6. $sum_n=b^infty Cn·X^n$ 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

share|cite|improve this question

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 | 

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) = $sum Dn·X^n$

1. f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$




2. f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$




3. $sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$




4. if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$




5. if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$




6. $sum_n=b^infty Cn·X^n$ 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

share|cite|improve this question

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) = $sum Dn·X^n$

1. f(x) ± g(x) = $sum_n=b^infty Cn·X^n$ ± $sum_n=b^infty Dn·X^n$ = $sum_n=b^infty (Cn ± Dn)·X^n$




2. f $(K · X^m)$ = $sum_n=b^infty Cn(K·X^m)^n$




3. $sum_n=b^infty Cn X^n+m $ = $X^m sum_n=b^infty Cn·X^n$




4. if f(x) = $sum_n=0^infty Cn·X^n$ then f'(x) = $sum_n=1^infty Cn·nX^n-1$




5. if f(x) = $sum_n=0^infty Cn·X^n$ then $int f(x)$ = $sum_n=0^infty$ $frac Cn·X^n+1 n+1$




6. $sum_n=b^infty Cn·X^n$ 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

share|cite|improve this question

asked Jul 30 at 5:34

LeviRivai

61

share|cite|improve this question

share|cite|improve this question

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 | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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 | 

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.

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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
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$

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answered Jul 30 at 5:57

Claude Leibovici

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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.

share|cite|improve this answer

answered Jul 30 at 5:42

copper.hat

122k557156

add a comment | 

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.

share|cite|improve this answer

answered Jul 30 at 5:42

copper.hat

122k557156

add a comment | 

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.

share|cite|improve this answer

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.

share|cite|improve this answer

answered Jul 30 at 5:42

copper.hat

122k557156

share|cite|improve this answer

share|cite|improve this answer

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 | 

add a comment | 

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
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$

share|cite|improve this answer

answered Jul 30 at 5:57

Claude Leibovici

111k1055126

add a comment | 

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
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$

share|cite|improve this answer

answered Jul 30 at 5:57

Claude Leibovici

111k1055126

add a comment | 

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
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$

share|cite|improve this answer

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
$$S_k=x^2sum_n=1^k n(n-1) x^n-2+xsum_n=1^k n x^n-1=x^2left(sum_n=1^k x^n right)''+xleft(sum_n=1^k x^n right)'$$

share|cite|improve this answer

answered Jul 30 at 5:57

Claude Leibovici

111k1055126

share|cite|improve this answer

share|cite|improve this answer

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 | 

add a comment |