Verified Solution • 2025–26 Session moderate

QJEEM20Maeab5: JEE Main 2020 Mathematics - Class 11 JEEM20Maeab5 Solution: Consider the data on x taking the values 0, 2, 4, 8,....., 2 n…

Consider the data on x taking the values
0, 2, 4, 8,....., 2n with frequencies
nC0 , nC1 , nC2 ,...., nCn respectively. If the
mean of this data is 7282n{{728} \over {{2^n}}} , then n is equal to _________ .

Step-by-Step Verified Solution

Mean = ∑x1.f1∑f1{{\sum {{x_1}.{f_1}} } \over {\sum {{f_1}} }}

= 0.nC0+2.nC1+22.nC2+...+2n.nCnnC0+nC1+...+nCn{{0.{}^n{C_0} + 2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}} \over {{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}}

We know,

(1 + x)n = nC0+nC1x+nC2x2+...+nCnxn{{}^n{C_0} + {}^n{C_1}x + {}^n{C_2}{x^2} + ... + {}^n{C_n}{x^n}} ...(1)

Put x = 2, at (1) we get
⇒\Rightarrow 3n - 1 = 2.nC1+22.nC2+...+2n.nCn{2.{}^n{C_1} + {2^2}.{}^n{C_2} + ... + {2^n}.{}^n{C_n}}

And Putting x = 1 in (1), we get

2n = nC0+nC1+...+nCn{{}^n{C_0} + {}^n{C_1} + ... + {}^n{C_n}}

∴\therefore Mean = 3n−12n{{{3^n} - 1} \over {{2^n}}}

According to question,

3n−12n{{{3^n} - 1} \over {{2^n}}} = 7282n{{728} \over {{2^n}}}

⇒\Rightarrow 3n = 729

⇒\Rightarrow n = 6

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