What will be the $n^{th}$ term of this given series? Given series is:
$$1+\frac{1\times x^2}{2\times 4}+\frac{1\times 3\times 5\times x^4}{2\times 4\times 6\times 8}+\frac{1\times 3\times 5\times 7\times 9\times x^6}{2\times 4\times 6\times 8\times 10\times 12}+.....\infty$$
I need to find it's $n^{th}$ term but am having trouble in dealing with increasing number of multiples in each step. Kindly Guide in How to proceed
 A: Hint: Your $n$th term (counting the first non-$1$ term as the first term and excepting the $0$th term) looks to be
$$x^{2n}\frac{1\times 3 \times \cdots \times  (4n-3)}{2\times 4\times \cdots \times (4n)}.$$
This equals
$$x^{2n}\frac{1\times 2 \times \cdots \times  (4n-2)}{\left(2\times 4\times \cdots \times (4n-2)\right)\left(2\times 4\times \cdots \times (4n)\right)}.$$
How can we represent this in terms of factorials?
A: $$a_0=1, \quad \text{and }a_n=\frac{x^{2n}}{2}\left(\prod_{k=0}^{n-1} \frac{2k+1}{2k+4}\right) \text{ for } n\ge1.$$
A: The general term will have, first of all, $x^{2n}$ in it, since we're looking at even powers of $x$. The denominator, in each case, is $2^{2n}(2n)!$. You can see this by separating out a factor of two from each multiplicand. The numerator of the coefficient appears to be $\frac{(4n-2)!}{2^{2n-1}(2n-1)!}$, but only for $n\geq 1$. Putting that all together, we have:
$$a_n=\frac{(4n-2)!}{2^{2n-1}(2n-1)!2^{2n}(2n)!}x^{2n} = \frac{(4n-2)!}{2^{4n-1}(2n)!(2n-1)!}x^{2n}, n\geq 1$$
To write out the entire series, you can use:
$$1+\sum_{n=1}^\infty \frac{(4n-2)!\,x^{2n}}{2^{4n-1}(2n)!(2n-1)!}$$
A: Let me expand on Carl's idea. Recall that $n!=1\times2\times3\times\cdots (n-1)\times n$. The product of the first $n$ even numbers, written as $2\times4\times6\times\cdots 2n$, can be expressed in terms of factorials: $2\times4\cdots 2n=2^n(1\times2\times3\cdots n)=2^n\times n!$ by factoring out a $2$ from each term.
The product of the first $n$ odd numbers, $1\times3\times5\times\cdots (2n-3)(2n-1)$, can also be written in terms of factorials: if $e_n$ is the product of the first $n$ even numbers, then $(2n)!/e_n=(1\times2\times3\cdots 2n)/(2\times4\times6\cdots 2n)=1\times3\times5\cdots(2n-1)$, so if $o_n$ is the product of the first $n$ odd numbers, then $o_n=(2n!)/e_n$.
With these two expressions, you can find your answer.
A: The correct power of $X$ for when $n$ is the $n^{th}$ term of the series is 
$$ a_n=K \times x^{2(n-1)} $$
You can deduce the correct formula starting from that
then for $ n>0 $ :
$$ a_n = \frac { K \times x^{2(n-1)} }{ 2^{2(n-1)} (2(n-1))!}  $$
For $ n > 1 $ :
$$ k = \frac {(4n-7)!}{2^{2(n-2)} (2(n-2))!}$$
Therefore,

$$ a_1 =1 $$
  For $n > 1$ :
  $$a_n = \frac { (4n-7)! \times x^{2(n-1)} }{ 2^{2(2n-3)} (2(n-2))! (2(n-1))!}  $$

