The mean of the coefficients of $x, x^{2}, \ldots, x^{7}$ in the binomial expansion of $(2+x)^{9}$ i

Question

The mean of the coefficients of $x, x^{2}, \ldots, x^{7}$ in the binomial expansion of $(2+x)^{9}$ is

Professor PC Thomas Classes · Answer

(2736) : $(2+x)^{9}={ }^{9} C_{0} 2^{9}+{ }^{9} C_{1} 2^{8} x^{1}$ $+{ }^{9} C_{2} 2^{7} x^{2}+{ }^{9} C_{3} 2^{6} x^{3}+{ }^{9} C_{4} 2^{5} x^{4}+{ }^{9} C_{5} 2^{4} x^{5}$\ $+{ }^{9} C_{6} 2^{3} x^{6}+{ }^{9} C_{7} 2^{2} x^{7}+{ }^{9} C_{8} 2 x^{8}+{ }^{9} C_{9} x^{9}$\ Coefficient of $x={ }^{9} C_{1} 2^{8}$\ Coeff. of $x^{2}={ }^{9} C_{2} 2^{7}, \ldots$, coeff. of $x^{7}={ }^{9} C_{7} 2^{2}$\ Mean $=\frac{{ }^{9} C_{1} \cdot 2^{8}+{ }^{9} C_{2} \cdot 2^{7} \ldots .+{ }^{9} C_{7} \cdot 2^{2}}{7}$\ $=\frac{(1+2)^{9}-{ }^{9} C_{0} \cdot 2^{9}-{ }^{9} C_{8} \cdot 2^{1}-{ }^{9} C_{9}}{7}$\ $=\frac{3^{9}-2^{9}-18-1}{7}=\frac{19152}{7}=2736$