$2 \cdot {}^{20}C_0 + 5 \cdot {}^{20}C_1 + 8 \cdot {}^{20}C_2 + 11 \cdot {}^{20}C_3 + \dots + 62 \cdot {}^{20}C_{20}$ is equal to

If ${ }^{n} C_0+\frac{1}{2}{ }^{n} C_1+\frac{1}{3}{ }^{n} C_2+\ldots+\frac{1}{n+1}{ }^{n} C_{n}=\frac{1023}{10}$,then $n=$

Statement $-1$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} = (n+2) 2^{n-1}$
Statement $-2$: $\sum_{r=0}^{n} (r+1) \binom{n}{r} x^r = (1+x)^n + nx(1+x)^{n-1}$

Let $a_0, a_1, a_2, \ldots, a_n \in \mathbb{R}$ be in an arithmetic progression and let $C_0, C_1, C_2, \ldots, C_n$ be the binomial coefficients. Then $\sum_{k=0}^n a_k \cdot C_k =$

$\sum\limits_{k = 0}^{10} {^{20}{C_k} = }$

$\frac{C_0}{1} + \frac{C_1}{2} + \frac{C_2}{3} + .... + \frac{C_n}{n + 1} = $