QJEEM23Ma8a49: JEE Main 2023 Mathematics - Class 11 JEEM23Ma8a49 Solution: The mean of the coefficients of in the binomial expansion of is …

We have, binomial coefficient, $(2+x)^9$

$T_{r+1}={ }^n C_r 2^{n-r} \times x^r$

Coefficient of $x\left(T_1\right)={ }^9 C_1 \times 2^8$

Coefficient of $x^2\left(T_2\right)={ }^9 C_2 \times 2^7$

Coefficient of $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
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Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$

$\text { Mean }=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7}$

$\begin{aligned}& { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots .+{ }^9 C_7 \times 2^2 \\& =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7} \end{aligned}$

$=\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736$