${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ is

Statement $-1$: $\mathop \sum \limits_{r = 0}^n \left( {r + 1} \right)\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right) = \left( {n + 2} \right){2^{n - 1}}$

Statement $-2$:$\;\mathop \sum \limits_{r = 0}^n \left( {r + 1} \right)\left( {\begin{array}{*{20}{c}}n\\r\end{array}} \right){x^r}\; = {\left( {1 + x} \right)^n} + nx{\left( {1 + x} \right)^{n - 1}}$

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$, then ${C_0}{C_2} + {C_1}{C_3} + {C_2}{C_4} + {C_{n - 2}}{C_n}$ equals

If number of terms in the expansion of ${(x - 2y + 3z)^n}$ are $45$, then $n=$

A possible value of $^{\prime}x^{\prime}$, for which the ninth term in the expansion of $\left\{3^{\log _{3} \sqrt{25^{x-1}+7}}+3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}\right\}^{10}$ in the increasing powers of $3^{\left(-\frac{1}{8}\right) \log _{3}\left(5^{x-1}+1\right)}$ is equal to $180$ , 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 :