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 ______.

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Statistics

Consider the data on

x

taking the values

0, 2, 4, 8, \dots, 2^{n}

with frequencies

^{n} C_{0},^{n} C_{1},^{n} C_{2}, \dots,^{n} C_{n}

respectively. If the mean of this data is

\frac{728}{2 ^{n}}

, then

n

is equal to ______.

Answer: 6

Order the data as the values

0, 2^{1}, 2^{2}, \dots, 2^{n}

carrying frequencies

(0 n), (1 n), \dots, (n n)

. The total frequency is

\sum_{k = 0}^{n} (k n) = 2^{n}

. For the numerator,

\sum_{k = 0}^{n} (k n) 2^{k} = (1 + 2)^{n} = 3^{n}

by the binomial theorem, but the first value is

0

rather than

2^{0} = 1

, so the

k = 0

term contributes nothing and the numerator is

3^{n} - 1

. The mean is therefore

\frac{3 ^{n} - 1}{2 ^{n}}

, and setting this equal to

\frac{728}{2 ^{n}}

gives

3^{n} - 1 = 728

, so

3^{n} = 729 = 3^{6}

and

n = 6

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Integer answer