If $1+\left(2+{ }^{49} C _{1}+{ }^{49} C _{2}+\ldots .+{ }^{49} C _{49}\right)\left({ }^{50} C _{2}+{ }^{50} C _{4}+\right.$ $\ldots . .+{ }^{50} C _{ so }$ ) is equal to $2^{ n } . m$, where $m$ is odd, then $n$ $+m$ is equal to.

If ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .... + {C_n}{x^n}$, then the value of ${C_0} + 2{C_1} + 3{C_2} + .... + (n + 1){C_n}$ will be

If $^{20}{C_1} + \left( {{2^2}} \right){\,^{20}}{C_3} + \left( {{3^2}} \right){\,^{20}}{C_3} + \left( {{2^2}} \right) + ..... + \left( {{{20}^2}} \right){\,^{20}}{C_{20}} = A\left( {{2^\beta }} \right)$, then the ordered pair $(A, \beta )$ is equal to

The sum of the last eight 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

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