Find the coefficient of $x^{29}$ in the given polynomial. 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.
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.
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
$$(x+0)(x+0)(x+0)...(x+0)(a_{30})+\\(x+0)(x+0)(x+0)...(a_{29})(x+0)+\\(x)(x)(x)...(a_{28})(x+0)(x+0)+\\
(x+0)(x+0)(x+0))...(a_{27})(x+0)(x+0)(x+0)+\\\vdots\\
(x+0)(a_{2})...(x+0)(x+0)(x+0)+\\
(x+a_1)(a+0)...(x+0)(x+0)(x+0)$$