Consider the data on $x$ taking the values $0,2,4,8 \ldots, 2^{n}$ with frequencies ${ }^{n} C_{0},{

Question

Consider the data on $x$ taking the values $0,2,4,8 \ldots, 2^{n}$ with frequencies ${ }^{n} C_{0},{ }^{n} C_{1},{ }^{n} C_{2}, \ldots,{ }^{n} C_{n}$ respectively. If the mean of this data is $\frac{728}{2^{n}}$, then $n$ is equal to

Professor PC Thomas Classes · Answer

(6) : We have the following information \Clearly, mean $$ \begin{aligned} & \begin{array}{r} 0 	imes{ }^{n} C_{0}+2 	imes{ }^{n} C_{1}+4 	imes{ }^{n} C_{2}+8 	imes{ }^{n} C_{3} \ +\ldots \ldots+2^{n} 	imes{ }^{n} C_{n} \end{array} \ = & \frac{\left[\begin{array}{r} { }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots . .+{ }^{n} C_{n} \ +\ldots .+{ }^{n} C_{n} \cdot 2^{n} \end{array}ight]}{2^{n}} \end{aligned} $$ $$ \left[\because{ }^{n} C_{0}+{ }^{n} C_{1}+{ }^{n} C_{2}+\ldots . .+{ }^{n} C_{n}=2^{n}ight] $$ $=\frac{(1+2)^{n}-1}{2^{n}}$ [Using Binomial theorem]\ $=\frac{3^{n}-1}{2^{n}}=\frac{728}{2^{n}}$\ (Given)\ $\Rightarrow 3^{n}-1=728 \Rightarrow 3^{n}=729=3^{6} \Rightarrow n=6$\