Question

Let $m, n \in N$ and $gcd (2, n) =1$ If
$30\begin{pmatrix}30\\ 0\end{pmatrix}+29 \begin{pmatrix}30\\ 1\end{pmatrix}+\ldots+2\begin{pmatrix}30\\ 28\end{pmatrix}+1\begin{pmatrix}30\\ 29\end{pmatrix}=n.2^{m}$, then n + m is equal to (Here $\begin{pmatrix}n\\ k\end{pmatrix}= ^{n}C_{k})$

Answer 1

$30\left({ }^{30} C _{0}\right)+29\left({ }^{30} C _{1}\right)+\ldots+2\left({ }^{30} C _{28}\right)+1\left({ }^{30} C _{29}\right)$
$=30\left({ }^{30} C _{30}\right)+29\left({ }^{30} C _{29}\right)+\ldots \ldots+2\left({ }^{30} C _{2}\right)+1\left({ }^{30} C _{1}\right)$
$=\displaystyle\sum_{ r =1}^{30} r \left({ }^{30} C _{ r }\right)$
$=\displaystyle\sum_{ r =1}^{30} r \left(\frac{30}{ r }\right)\left({ }^{29} C _{ r -1}\right)$
$=30 \displaystyle\sum_{ r =1}^{30}{ }^{29} C _{ r -1}$
$=30\left({ }^{29} C _{0}+{ }^{29} C _{1}+{ }^{29} C _{2}+\ldots+{ }^{29} C _{29}\right)$
$=30\left(2^{29}\right)=15(2)^{30}= n (2)^{ m }$
$\therefore n =15, m =30$
$n + m =45$