If the sum of the coefficients of all even powers of $x$ in the product $\left(1+x+x^{2}+\ldots+x^{2 n}\right)\left(1-x+x^{2}-x^{3}+\ldots+x^{2 n}\right)$ is $61,$ then $\mathrm{n}$ is equal to

If ${C_0},{C_1},{C_2},.......,{C_n}$ are the binomial coefficients, then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals

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

The sum of last eigth coefficients in the expansion of $(1 + x)^{15}$ is :-

Let $\left( a + bx + cx ^2\right)^{10}=\sum \limits_{ i =0}^{20} p _{ i } x ^{ i }, a , b , c \in N$. If $p _1=20$ and $p _2=210$, then $2( a + b + c )$ is equal to

The value of ${\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}\left( { - 1} \right)}}{{{2^{2r + 1}}}}} ^r}$ is