Question

Let $m,n\in N$ and $gcd(2,n)=1$. If $30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=n.2^m$, then $n+m$ is equal to

Answer 1

Finding the value of $n+m$

Given, $gcd\left(2,n\right)=1$ and $30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=n.2^m$

We know that,

$$\begin{gathered} 30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=\sum _{r=1}^{30}r\left({}^{30}C_r\right) \\ \Rightarrow 30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=\sum _{r=1}^{30}r\left(\frac{30}{r}\right){}^{29}C_{r-1}\left[\because {}^{n}C_r=\frac{n}{r}\left({}^{n-1}C_{r-1}\right)\right] \\ \Rightarrow 30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=30\left({}^{29}C_0+{}^{29}C_1+....+{}^{29}C_{29}\right)\left[\text{Putting the value of}r\right] \\ \Rightarrow 30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=30\times 2^{29}\left[\therefore {}^{n}C_0+{}^{n}C_1+{}^{n}C_2+{}^{n}C_3...{}^{n}C_n=2^n\right] \\ \Rightarrow 30{}^{30}C_0+29{}^{30}C_1+\dots ..+2{}^{30}C_{28}+1{}^{30}C_{29}=15\times 2^{30} \end{gathered}$$

Comparing the value, we get, $n=15$ and $m=30$.

Therefore, the value of $n+m=15+30=45$