The sum of the coefficients of even power of $x$ in the expansion of ${(1 + x + {x^2} + {x^3})^5}$ is

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+ x )^{99}$. Let a be the middle term in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\frac{{ }^{200} C _{99} K }{ a }=\frac{2^{\ell} m }{ n }$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n )$ is equal to :

If $\mathrm{b}$ is very small as compared to the value of $\mathrm{a}$, so that the cube and other higher powers of $\frac{b}{a}$ can be neglected in the identity $\frac{1}{a-b}+\frac{1}{a-2 b}+\frac{1}{a-3 b} \ldots .+\frac{1}{a-n b}=\alpha n+\beta n^{2}+\gamma n^{3}$, then the value of $\gamma$ is:

If the sum of the coefficients in the expansion of $(x+y)^{n}$ is $4096,$ then the greatest coefficient in the expansion is .... .

Value $\sum\limits_{r = 0}^{15} {\left( {{}^{15}{C_r}{}^{40}{C_{15}}{}^{20}{C_r} - {}^{35}{C_{15}}{}^{15}{C_r}{}^{25}{C_r}} \right)} $ is-

The coefficient of $x ^{101}$ in the expression $(5+x)^{500}+x(5+x)^{499}+x^{2}(5+x)^{498}+\ldots . x^{500}$ $x>0$, is