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The polynomial is :

$$ \left(x-\frac{1}{1\cdot3}\right) \left(x-\frac{2}{1\cdot3\cdot5}\right) \left(x-\frac{3}{1\cdot3\cdot5\cdot7}\right) \cdots \left(x-\frac{30}{1\cdot3\cdot5\cdots61}\right) $$

What I've done so far : The given polynomial is an expression of degree $30$. Hence, the coefficient of $x^{29}$ will be the negative of the sum of the roots. But the resulting sum is too complicated to handle and I think I'm doing it wrong.

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    No, it's minus the sum of the roots. What you've described is, up to sign, the coefficient of $x$. – 2017-02-19
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    Oh... So the answer is the negative of the sum of the roots ? – 2017-02-19
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    If there are $30$ terms $(x-a_i)$, then the only way to get $x^{29}$ is by choosing $29$ $x$'s and a single $-a_i$. So the coefficient of $x^{29}$ is going to be $-\sum_{i=1}^{30}a_i$. – 2017-02-19
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    The sum of the roots can be guessed by examining the sum of the first $1, 2, 3, 4$ terms and then proved by induction. – 2017-02-19
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    @user49640 Isn't there a way to do it using the binomial theorem ? – 2017-02-19
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    I'm not sure.${}$ – 2017-02-19
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    See [Vieta's Formulas](https://en.wikipedia.org/wiki/Vieta's_formulas) – 2017-02-19
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    @Dr.MV Vieta's Formulas have helped me find that the coefficient is actually the sum of the roots of the equation. How do I find the sum of the roots ? – 2017-02-19
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    I'm looking for an answer in terms of $$ 1.3.5.....61 $$ – 2017-02-19
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    1/3=1/3;1/3+2/15=7/15;1/3+2/15+3/105=52/105; I'm seeing a pattern and a reason. – 2017-02-19

4 Answers 4

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HINT: The sum $S_k $ given by

$$\sum_{n=1}^{k} \dfrac {n}{(2n+1)!!} $$

for $k=1,2,3....$ gives

$$S_1=\frac {1}{3} \\ =\frac {1}{2}-\frac {1/2}{ 1 \cdot 3}$$

$$S_2=\frac {1}{3} + \frac {2}{15} =\frac {7}{15} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 }$$

$$S_3=\frac {1}{3} + \frac {2}{15} +\frac {3}{105} =\frac {52}{105} \\= \frac {1}{2}-\frac {1/2}{1 \cdot 3 \cdot 5 \cdot 7 }$$

You can use induction to confirm this pattern.

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The coefficient is $-\sum_{i=1}^{30}\frac {i}{\prod_{j=1}^i (2j+1)} $

Claim: let $M_n = \prod_{j=1}^n (2j+1) $ then $\sum_{i=1}^{n}\frac {i}{\prod_{j=1}^i (2j+1)}=\frac {M_n-1}{2M_n} $.

Proof by induction:

Base step: $1/1*3 = \frac {3-1}{2*3} $.

Induction step: Suppose $\sum_{i=1}^{n}\frac {i}{\prod_{j=1}^i (2j+1)}=\frac {M_n-1}{2M_n} = \frac {M_n-1}{2M_n}$. Then for $n+1$:

$\frac {M_n-1}{2M_n}+\frac {n+1}{M_n*(2(n+1)+1)}=\frac {(M_n-1)(2 (n+1)+1)}{2M_n*(2 (n+1)+1)}+\frac { (n+1)*2}{2M_{n+1}} $

$= \frac {(M_n-1)(2 (n+1)+1)+ 2(n+1)}{2M_{n+1}}=$

$\frac {M_{n+1}-2 (n+1)-1+2(n+1)}{M_{n+1}}=$

$\frac {M_{n+1}-1}{M_{n+1}}=$

So the coefficient is:

$-\frac {1*3*5*7*....*61-1}{2(1*3*5*7*....*61)} $

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Let $p_m(x) =\prod_{k=1}^{m} (x-\dfrac{k}{\prod_{j=1}^{k+1}(2j-1)}) $

Since, in the usual way,

$\begin{array}\\ \prod_{j=1}^{k+1}(2j-1) &=\dfrac{\prod_{j=1}^{2k+1}j}{\prod_{j=1}^k(2j)}\\ &=\dfrac{(2k+1)!}{2^kk!}\\ \end{array} $

$p_m(x) =\prod_{k=1}^{m} (x-\dfrac{k2^kk!}{(2k+1)!}) $

The coefficient of $x^{m-1}$ is $(-1)^{m-1}$ times the sum of the roots, which is

$\begin{array}\\ \sum_{k=1}^m \dfrac{k}{\dfrac{(2k+1)!}{2^kk!}} &=\sum_{k=1}^m \dfrac{k2^kk!}{(2k+1)!} \\ \end{array} $

According to Wolfy, this sum approaches $\frac12$ as $m \to \infty$.

Calling this sum $s(m)$, then $s(2:5) =(7/15, 52/105, 472/945, 5197/10395) $.

To see how close this is to $\frac12$, $\frac12-s(2:5) =(1/35, 1/210, 1/1890, 1/20790) $.

For $m=30$, $\frac12-s(30) =\dfrac1{3564303977319726652772203331132409634218750} $.

Note that $s(m) =\sum_{k=1}^m \dfrac{k2^kk!}{(2k+1)!} =\dfrac1{(2m+1)!}\sum_{k=1}^m k2^kk!\dfrac{(2m+1)!}{(2k+1)!} $ which explains, to a certain extent, that denominator. Im particular, all its prime factors do not exceed $2m+1$.

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HINT:$$\underbrace{1.3,1.3.5,...,1.3.5....61}_{30-terms} $$so $$\underbrace{(x-\frac{1}{1.3})(x-\frac{2}{1.3.5})(x-\frac{3}{1.3.5.7})......(x-\frac{30}{1.3.5.....61})}_{30-terms }$$we have ,to obtain $x^{29}$ we have when you want $x^{29}$ from $(x+a_1)(x+a_2)(x+a_3)...(x+a_{30})$ you have to find this
$$(x+0)(x+0)(x+0)...(x+0)(a_{30})+\\(x+0)(x+0)(x+0)...(a_{29})(x+0)+\\(x)(x)(x)...(a_{28})(x+0)(x+0)+\\ (x+0)(x+0)(x+0))...(a_{27})(x+0)(x+0)(x+0)+\\\vdots\\ (x+0)(a_{2})...(x+0)(x+0)(x+0)+\\ (x+a_1)(a+0)...(x+0)(x+0)(x+0)$$

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    The op has stated he has done that but doesn't know the sum. – 2017-02-19
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    I'll correct it – 2017-02-19