Find the coefficient of x^29 in the expansion of a series

[cooldudeachyut]

Question - Find the coefficient of x²⁹ in the expansion of (x-(1/(1*3)))(x-(2/(1*3*5)))(x-(3/(1*3*5*7))).......(x-(30/(1*3*5......61)))

My attempt : I tried to write the series as :

[FONT=MathJax_Size1]_k=1∏^k=30(x-(k/(1*3*5......2k+1)))
[/FONT]
= _k=1[FONT=MathJax_Size1]∏^k=30[/FONT][FONT=MathJax_Size1](x-(1/2)(2k+1-1/(1*3*5......2k+1)))[/FONT][FONT=MathJax_Size1]

= [/FONT]_k=1[FONT=MathJax_Size1]∏^k=30[/FONT][FONT=MathJax_Size1](x+(1/2)(1/(1*3*5......2k+1) - 1/(1*3*5......2k-1))) {May help probably}

[/FONT][FONT=MathJax_Size1]After that I cannot think of anything on how to proceed to solve this problem. Please help.
[/FONT][FONT=MathJax_Size1]
[/FONT]

 

[Ishuda]

Question - Find the coefficient of x²⁹ in the expansion of (x-(1/(1*3)))(x-(2/(1*3*5)))(x-(3/(1*3*5*7))).......(x-(30/(1*3*5......61)))

My attempt : I tried to write the series as :

[FONT=MathJax_Size1]_k=1∏^k=30(x-(k/(1*3*5......2k+1)))
[/FONT]
= _k=1[FONT=MathJax_Size1]∏^k=30[/FONT][FONT=MathJax_Size1](x-(1/2)(2k+1-1/(1*3*5......2k+1)))[/FONT][FONT=MathJax_Size1]

= [/FONT]_k=1[FONT=MathJax_Size1]∏^k=30[/FONT][FONT=MathJax_Size1](x+(1/2)(1/(1*3*5......2k+1) - 1/(1*3*5......2k-1))) {May help probably}


[/FONT][FONT=MathJax_Size1]After that I cannot think of anything on how to proceed to solve this problem. Please help.
[/FONT][FONT=MathJax_Size1]
[/FONT]

Or, look at
http://www.salonhogar.net/themathpage/aPreCalc/proof-binomial-theorem.htm

If we let
c_k = \(\displaystyle \frac{k}{1 \cdot 3 \cdot 5 \cdot ... \cdot (2k-1) \cdot (2k+1)}\)
and label the expression f(x), then we can write
f(x) = \(\displaystyle \Pi_{k=1}^{k=30}\, (x\, -\, c_k)\)

Now, writing f(x) as
f(x) = \(\displaystyle \Sigma_{k=0}^{k=30}\, a_k\, x^k\),
the above link implies (and it is easily proven by induction)
\(\displaystyle a_{29}\, =\, \Sigma_{k=1}^{k=30}\, c_k\)

EDIT: Fix c_k definition - thnx (possible) decedent of the great Ghengis

 

[Deleted member 4993]

EDIT: Fix c_k definition - thnx (possible) decedent of the great Ghengis

Really??!!!

- you know something I don't know....

Decedent

An individual who has died. The term literally means "one who is dying," but it is commonly used in the law to denote one who has died, particularly someone who has recently passed away.

 

[Ishuda]

Really??!!!

- you know something I don't know....

Decedent

An individual who has died. The term literally means "one who is dying," but it is commonly used in the law to denote one who has died, particularly someone who has recently passed away.

Once again I have let the computer choose a word for me despite it having been demonstrated many many times that doing so is a very foolish action. It has been said that to apologize in the Japanese culture means that you take the responsibility for what has happened and acknowledge that the action was under your control [thus to apologize to a superior is very incorrect since it means you have control of their life, at least in some small part]. Therefore, since I greatly admire the Japanese culture, I will not apologize but rather abase myself and only say I now know what I did was an incorrect and foolish action and hope for a chance to show my appreciation for your continued guidance. Or, if not, may I at least have the honor of committing seppuku because of the shame I have brought on myself?

 

[Deleted member 4993]

.... may I at least have the honor of committing seppuku because of the shame I have brought on myself?

Seppuku ... I did not you were a Samurai ...

But I think that would be too easy - I have to consult my great-great-great-great-great-great-great-grandfather for proper punishment ....

 

[Ishuda]

Seppuku ... I did not you were a Samurai ...

But I think that would be too easy - I have to consult my great-great-great-great-great-great-great-grandfather for proper punishment ....

I humbly await you most welcome instructions.

 

[cooldudeachyut]

Or, look at
http://www.salonhogar.net/themathpage/aPreCalc/proof-binomial-theorem.htm

If we let
c_k = \(\displaystyle \frac{k}{1 \cdot 3 \cdot 5 \cdot ... \cdot (2k-1) \cdot (2k+1)}\)
and label the expression f(x), then we can write
f(x) = \(\displaystyle \Pi_{k=1}^{k=30}\, (x\, -\, c_k)\)

Now, writing f(x) as
f(x) = \(\displaystyle \Sigma_{k=0}^{k=30}\, a_k\, x^k\),
the above link implies (and it is easily proven by induction)
\(\displaystyle a_{29}\, =\, \Sigma_{k=1}^{k=30}\, c_k\)

EDIT: Fix c_k definition - thnx (possible) decedent of the great Ghengis

Thanks