<s>[INST] 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
 [/INST] 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 $n > 1$ :
  $$a_n = \frac { (4n-7)! \times x^{2(n-1)} }{ 2^{2(2n-3)} (2(n-2))! (2(n-1))!}  $$

</s>