The mean of the coefficients of x, x^2, ldots | vedclass

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(B) The binomial expansion is $(2+x)^9 = \sum_{r=0}^{9} {^9C_r} \cdot 2^{9-r} \cdot x^r$.The coefficient of $x^r$ is $T_r = {^9C_r} \cdot 2^{9-r}$.We

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$2736$

Vedclass · Answer

(B) The binomial expansion is $(2+x)^9 = \sum_{r=0}^{9} {^9C_r} \cdot 2^{9-r} \cdot x^r$.The coefficient of $x^r$ is $T_r = {^9C_r} \cdot 2^{9-r}$.We need the mean of the coefficients of $x, x^2, \ldots, x^7$,which is $S = \frac{1}{7} \sum_{r=1}^{7} {^9C_r} \cdot 2^{9-r}$.We know that $\sum_{r=0}^{9} {^9C_r} \cdot 2^{9-r} = (2+1)^9 = 3^9 = 19683$.Thus,$\sum_{r=1}^{7} {^9C_r} \cdot 2^{9-r} = 3^9 - ({^9C_0} \cdot 2^9 + {^9C_8} \cdot 2^1 + {^9C_9} \cdot 2^0)$.Calculating the terms: ${^9C_0} \cdot 2^9 = 1 \cdot 512 = 512$,${^9C_8} \cdot 2^1 = 9 \cdot 2 = 18$,and ${^9C_9} \cdot 2^0 = 1 \cdot 1 = 1$.Sum $= 19683 - (512 + 18 + 1) = 19683 - 531 = 19152$.Mean $= \frac{19152}{7} = 2736$.

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

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