The sum of all the coefficients in the binomial expansion of $(1+2x)^n$ is $6561$. Let $R=(1+2x)^n=I+F$,where $I \in N$ and $0 < F < 1$. If $x=\frac{1}{\sqrt{2}}$,then $1-\frac{F}{1+(\sqrt{2}-1)^4}=$

Let $(1+x+x^2)^{2014} = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + \ldots + a_{4028} x^{4028}$. Let $A = a_0 - a_3 + a_6 - \ldots + a_{4026}$,$B = a_1 - a_4 + a_7 - \ldots - a_{4027}$,and $C = a_2 - a_5 + a_8 - \ldots + a_{4028}$. Then,

Let $\lambda$ be the positive root of the equation $x^2-x-1=0$,and set $a_n = \frac{1}{\sqrt{5}}\left(\lambda^n - (1-\lambda)^n\right)$ for $n \in N$,where $N$ is the set of all natural numbers. Consider the sets $A = \{ n \in N : a_n \text{ is a rational number, but not an integer} \}$ and $B = \{ n \in N : a_n \text{ is an irrational number} \}$. Then:

Let $M = 2^{30} - 2^{15} + 1$. When $M^2$ is expressed in base $2$,the number of $1$'s in its binary representation is:

Let $(1 + x + x^2)^{20}(2x + 1) = a_0 + a_1x^1 + a_2x^2 + ... + a_{41}x^{41}$,then $\frac{a_0}{1} + \frac{a_1}{2} + .... + \frac{a_{41}}{42}$ is equal to

For the natural numbers $m, n$,if $(1-y)^{m}(1+y)^{n}=1+a_{1} y+a_{2} y^{2}+\ldots +a_{m+n} y^{m+n}$ and $a_{1}=a_{2}=10$,then the value of $(m+n)$ is equal to: