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Find the formula for the sum: $1+3x^2+5x^4+7x^6+…+(2n+1)x^2n$

Rearrange the formula for the sum of a geometric series to find the value of its common ratio?Formula for the sum of the following sequenceFinding the formula of a sequence.Given formula to calculate sum of first n terms of a sequence, show that the sequence is geometricFind the series: $frac-14+left(frac12+frac14+frac28+frac316+frac532+cdotsright)$Find the sum for the power seriesFormula for the nth partial sum of a telescoping seriesFind the common ratio given $n$ via sumApplying the Geometric Series Formula w/ This Real Life Situation?How to find the partial sum of $n/2^n$?

2

$begingroup$

The formula is supposed to be valid for $x neq pm 1$.

Here is how I did it:

$$barS_n = fracddxleft( x+x^3+x^5+...+x^2n+1 right)$$

The term in the brackets is the geometric sequence. So call the sum up to term $x^2n+1$ as $S_n$, then:

$$S_n = fracx(1-x^2n)1-x^2$$

and so:

$$barS_n = fracddx left( S_n right)$$

which gives me:

$$frac2x^2 (1-x^2n)(1-x^2)^2+frac11-x^2left( 1-x^2n(1+2n) right)$$

apparently, that's wrong. Embarrassing I cannot solve such a simple problem.

sequences-and-series derivatives

share|cite|improve this question

edited Mar 12 at 10:02

i squared - Keep it Real

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

add a comment | 

2

$begingroup$

The formula is supposed to be valid for $x neq pm 1$.

Here is how I did it:

$$barS_n = fracddxleft( x+x^3+x^5+...+x^2n+1 right)$$

The term in the brackets is the geometric sequence. So call the sum up to term $x^2n+1$ as $S_n$, then:

$$S_n = fracx(1-x^2n)1-x^2$$

and so:

$$barS_n = fracddx left( S_n right)$$

which gives me:

$$frac2x^2 (1-x^2n)(1-x^2)^2+frac11-x^2left( 1-x^2n(1+2n) right)$$

apparently, that's wrong. Embarrassing I cannot solve such a simple problem.

sequences-and-series derivatives

share|cite|improve this question

edited Mar 12 at 10:02

i squared - Keep it Real

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

add a comment | 

$begingroup$

The formula is supposed to be valid for $x neq pm 1$.

Here is how I did it:

$$barS_n = fracddxleft( x+x^3+x^5+...+x^2n+1 right)$$

The term in the brackets is the geometric sequence. So call the sum up to term $x^2n+1$ as $S_n$, then:

$$S_n = fracx(1-x^2n)1-x^2$$

and so:

$$barS_n = fracddx left( S_n right)$$

which gives me:

$$frac2x^2 (1-x^2n)(1-x^2)^2+frac11-x^2left( 1-x^2n(1+2n) right)$$

apparently, that's wrong. Embarrassing I cannot solve such a simple problem.

sequences-and-series derivatives

share|cite|improve this question

edited Mar 12 at 10:02

i squared - Keep it Real

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

$endgroup$

share|cite|improve this question

edited Mar 12 at 10:02

i squared - Keep it Real

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

share|cite|improve this question

edited Mar 12 at 10:02

i squared - Keep it Real

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

asked Mar 12 at 9:49

i squared - Keep it Reali squared - Keep it Real

1,61311027

3 Answers
3

active

oldest

votes

3

$begingroup$

There is a minor mistake.

$$S_n=x(1+x^2+x^4+cdots+x^2n)=fracx(1-x^2n+2)1-x^2$$

share|cite|improve this answer

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

$endgroup$

add a comment | 

1

$begingroup$

Your sum is given by $$fracx^2 left(-2 n x^2 n-3 x^2 n+2 n x^2
n+2+x^2 n+2-x^2+3right)left(x^2-1right)^2$$

share|cite|improve this answer

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So my mistake is in the differentiation step?
  $endgroup$
  – i squared - Keep it Real
  Mar 12 at 10:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think so,you can compare with the solution above.
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 12 at 10:06
  

  
  
  
  

add a comment | 

0

$begingroup$

Without calculus:

$$S_3:=1+3x^2+5x^4+7x^6
\=2(1+x^2+x^4+x^6)-1+x^2(1+3x^2+5x^4)
\=2frac1-x^81-x^2-1+x^2(S_3-7x^6)$$

and more generally

$$S_n=2frac1-x^2n+21-x^2-1+x^2(S_n-(2n+1)x^2n).$$

So

$$(1-x^2)S_n=2frac1-x^2n+21-x^2-1-(2n+1)x^2n+2.$$

share|cite|improve this answer

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

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3

$begingroup$

There is a minor mistake.

$$S_n=x(1+x^2+x^4+cdots+x^2n)=fracx(1-x^2n+2)1-x^2$$

share|cite|improve this answer

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

$endgroup$

add a comment | 

3

$begingroup$

There is a minor mistake.

$$S_n=x(1+x^2+x^4+cdots+x^2n)=fracx(1-x^2n+2)1-x^2$$

share|cite|improve this answer

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

$endgroup$

add a comment | 

$begingroup$

There is a minor mistake.

$$S_n=x(1+x^2+x^4+cdots+x^2n)=fracx(1-x^2n+2)1-x^2$$

share|cite|improve this answer

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

$endgroup$

share|cite|improve this answer

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

answered Mar 12 at 10:11

CY AriesCY Aries

16.8k11743

1

$begingroup$

Your sum is given by $$fracx^2 left(-2 n x^2 n-3 x^2 n+2 n x^2
n+2+x^2 n+2-x^2+3right)left(x^2-1right)^2$$

share|cite|improve this answer

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So my mistake is in the differentiation step?
  $endgroup$
  – i squared - Keep it Real
  Mar 12 at 10:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think so,you can compare with the solution above.
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 12 at 10:06
  

  
  
  
  

add a comment | 

1

$begingroup$

Your sum is given by $$fracx^2 left(-2 n x^2 n-3 x^2 n+2 n x^2
n+2+x^2 n+2-x^2+3right)left(x^2-1right)^2$$

share|cite|improve this answer

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  So my mistake is in the differentiation step?
  $endgroup$
  – i squared - Keep it Real
  Mar 12 at 10:03
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I think so,you can compare with the solution above.
  $endgroup$
  – Dr. Sonnhard Graubner
  Mar 12 at 10:06
  

  
  
  
  

add a comment | 

$begingroup$

Your sum is given by $$fracx^2 left(-2 n x^2 n-3 x^2 n+2 n x^2
n+2+x^2 n+2-x^2+3right)left(x^2-1right)^2$$

share|cite|improve this answer

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

$endgroup$

share|cite|improve this answer

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

answered Mar 12 at 9:55

Dr. Sonnhard GraubnerDr. Sonnhard Graubner

77.8k42866

0

$begingroup$

Without calculus:

$$S_3:=1+3x^2+5x^4+7x^6
\=2(1+x^2+x^4+x^6)-1+x^2(1+3x^2+5x^4)
\=2frac1-x^81-x^2-1+x^2(S_3-7x^6)$$

and more generally

$$S_n=2frac1-x^2n+21-x^2-1+x^2(S_n-(2n+1)x^2n).$$

So

$$(1-x^2)S_n=2frac1-x^2n+21-x^2-1-(2n+1)x^2n+2.$$

share|cite|improve this answer

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

0

$begingroup$

Without calculus:

$$S_3:=1+3x^2+5x^4+7x^6
\=2(1+x^2+x^4+x^6)-1+x^2(1+3x^2+5x^4)
\=2frac1-x^81-x^2-1+x^2(S_3-7x^6)$$

and more generally

$$S_n=2frac1-x^2n+21-x^2-1+x^2(S_n-(2n+1)x^2n).$$

So

$$(1-x^2)S_n=2frac1-x^2n+21-x^2-1-(2n+1)x^2n+2.$$

share|cite|improve this answer

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229

$endgroup$

add a comment | 

$begingroup$

Without calculus:

$$S_3:=1+3x^2+5x^4+7x^6
\=2(1+x^2+x^4+x^6)-1+x^2(1+3x^2+5x^4)
\=2frac1-x^81-x^2-1+x^2(S_3-7x^6)$$

and more generally

$$S_n=2frac1-x^2n+21-x^2-1+x^2(S_n-(2n+1)x^2n).$$

So

$$(1-x^2)S_n=2frac1-x^2n+21-x^2-1-(2n+1)x^2n+2.$$

share|cite|improve this answer

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229

$endgroup$

share|cite|improve this answer

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229

answered Mar 12 at 10:43

Yves DaoustYves Daoust

130k676229