2 for the cost of 1

This solution will help the Indian guys on this site as it tells the way which can be used to solve such kinds of problems quite easily and such questions are quite regularly asked in JEE.So,here goes, $$ In such problems,the best way to go is to find a relation between $A$ and $B$ .In this case,to find $A$ we have to think co-efficients and even in $B$ ,the given terms are binomial co-efficients.So,we have got a relation now let us build on it: ${\left(x+x^2+\dfrac{1}{x}+\dfrac{1}{x^2}\right)}^{15}\\ =\dfrac{{(x^4+x^3+x+1)}^{15}}{x^{30}}\\ =\dfrac{[{(x+1)(x^3+1)}]^{15}}{x^{30}}$ .Now,we just have to find the co-efficient of $x^{30}$ in the expansion of $[{(x+1)(x^3+1)}]^{15}$ .Now,let us assume that the $r^{th}$ term of the first expression $\times$ $q^{th}$ term of the second expression gives us $x^{30}$ ,then we can write it as, $\dbinom{15}{r}x^r\times\dbinom{15}{q}x^{3q}=x^{30}\times \text{the required co-efficient}\\ \Longrightarrow r+3q=30$ ,now,we list the solutions of the equation keeping in mind that $r$ has to be less than $15$ ,we get $(r,q)\equiv (15,5),(12,6),(9,7),(6,8),(3,9),(0,10)$ ,now we have to calculate the sum of the co-efficients which would be equal to $B$ ,yes!When you find the sum of co-efficients you will find that they are the same as the elements of $B$ ,hence, $\dfrac{2A}{B}=2$ .And done!