If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots+C_n x^n$,then $C_0+2 C_1+3 C_2+\ldots+(n+1) C_n$ is equal to

If ${}^{21}C_1 + 3 \cdot {}^{21}C_3 + 5 \cdot {}^{21}C_5 + \dots + 19 \cdot {}^{21}C_{19} + 21 \cdot {}^{21}C_{21} = k$,then the number of prime factors of $k$ is:

In the expansion of $(1 + x)^n$,the sum of the coefficients of odd powers of $x$ is:

The mean of the values $0, 1, 2, \dots, n$ having corresponding weights $^nC_0, ^nC_1, ^nC_2, \dots, ^nC_n$ respectively is

Evaluate the sum: $\left( \binom{21}{1} - \binom{10}{1} \right) + \left( \binom{21}{2} - \binom{10}{2} \right) + \left( \binom{21}{3} - \binom{10}{3} \right) + \dots + \left( \binom{21}{10} - \binom{10}{10} \right) = $

In the expansion of $(1 + x)^{50}$,the sum of the coefficients of odd powers of $x$ is