Question

Let $m,n\in N$ and $gcd(2,n)=1$ If
$30\left(\begin{array}{c}30\\ 0\end{array}\right)+29\left(\begin{array}{c}30\\ 1\end{array}\right)+\dots +2\left(\begin{array}{c}30\\ 28\end{array}\right)+1\left(\begin{array}{c}30\\ 29\end{array}\right)=n.2^m$, then n + m is equal to (Here $\left(\begin{array}{c}n\\ k\end{array}\right){=}^n{C}_k)$

Answer 1

Answer: 45

$30\left({}^{30}{C}_0\right)+29\left({}^{30}{C}_1\right)+\dots +2\left({}^{30}{C}_{28}\right)+1\left({}^{30}{C}_{29}\right)$
$=30\left({}^{30}{C}_{30}\right)+29\left({}^{30}{C}_{29}\right)+\dots \dots +2\left({}^{30}{C}_2\right)+1\left({}^{30}{C}_1\right)$
$=\sum _{r=1}^{30}r\left({}^{30}{C}_r\right)$
$=\sum _{r=1}^{30}r\left(\frac{30}{r}\right)\left({}^{29}{C}_{r-1}\right)$
$=30\sum _{r=1}^{30}{}^{29}{C}_{r-1}$
$=30\left({}^{29}{C}_0+{}^{29}{C}_1+{}^{29}{C}_2+\dots +{}^{29}{C}_{29}\right)$
$=30\left(2^{29}\right)=15(2{)}^{30}=n(2{)}^m$
$\therefore n=15,m=30$
$n+m=45$