JEE MAIN - Mathematics (2023 - 11th April Morning Shift - No. 22)
The mean of the coefficients of $$x, x^{2}, \ldots, x^{7}$$ in the binomial expansion of $$(2+x)^{9}$$ is ___________.
Answer
2736
Explanation
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 $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
. .
. .
. .
Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$
$$ \text { Mean }=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7} $$
$$ \begin{aligned} & { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots .+{ }^9 C_7 \times 2^2 \\ & =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7} \end{aligned} $$
$$ =\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736 $$
$$ T_{r+1}={ }^n C_r 2^{n-r} \times x^r $$
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 $x^3\left(T_3\right)={ }^9 C_3 \times 2^6$
. .
. .
. .
Coefficient of $x^7\left(T_7\right)={ }^9 C_7 \times 2^2$
$$ \text { Mean }=\frac{{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots+{ }^9 C_7 \times 2^2}{7} $$
$$ \begin{aligned} & { }^9 C_0 \times 2^9+{ }^9 C_1 \times 2^8+{ }^9 C_2 \times 2^7+{ }^9 C_3 \times 2^6+\ldots .+{ }^9 C_7 \times 2^2 \\ & =\frac{+{ }^9 C_8 \times 2^1+{ }^9 C_9 \times 2^0-{ }^9 C_0 \times 2^9-{ }^9 C_8 \times 2^1-{ }^9 C_9 \times 2^0}{7} \end{aligned} $$
$$ =\frac{(1+2)^9-2^9-18-1}{7}=\frac{19152}{7}=2736 $$
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