# Solve the Following Questions

Question:

Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If

$30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)=\mathrm{n} \cdot 2^{\mathrm{m}}$, then $\mathrm{n}+\mathrm{m}$ is equal to

(Here $\left(\begin{array}{l}\mathrm{n} \\ \mathrm{k}\end{array}\right)={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{k}}$ )

Solution:

$30\left({ }^{30} \mathrm{C}_{0}\right)+29\left({ }^{30} \mathrm{C}_{1}\right)+\ldots+2\left({ }^{30} \mathrm{C}_{28}\right)+1\left({ }^{30} \mathrm{C}_{29}\right)$

$=30\left({ }^{30} \mathrm{C}_{30}\right)+29\left({ }^{30} \mathrm{C}_{29}\right)+\ldots \ldots+2\left({ }^{30} \mathrm{C}_{2}\right)+1\left({ }^{30} \mathrm{C}_{1}\right)$

$=\sum_{\mathrm{r}=1}^{30} \mathrm{r}\left({ }^{30} \mathrm{C}_{\mathrm{r}}\right)$

$=\sum_{\mathrm{r}=1}^{30} \mathrm{r}\left(\frac{30}{\mathrm{r}}\right)\left({ }^{29} \mathrm{C}_{\mathrm{r}-1}\right)$

$=30 \sum_{\mathrm{r}=1}^{30}{ }^{29} \mathrm{C}_{\mathrm{r}-1}$

$=30\left({ }^{29} \mathrm{C}_{0}+{ }^{29} \mathrm{C}_{1}+{ }^{29} \mathrm{C}_{2}+\ldots+{ }^{29} \mathrm{C}_{29}\right)$

$=30\left(2^{29}\right)=15(2)^{30}=\mathrm{n}(2)^{\mathrm{m}}$

$\therefore \mathrm{n}=15, \mathrm{~m}=30$

$\mathrm{n}+\mathrm{m}=45$