Discussion

Let \((2 - x)^{9} = a_{0} + a_{1}x + a_{2}x^{2} + \cdots + a_{9}x^{9}\). Then the value of \(a_{1} + a_{2} + a_{3} + \cdots + a_{8}\) is equal to

Answer: E

The sum is \(-510\).

Put \(x = 1\) to get the sum of all coefficients, then remove \(a_{0}\) and \(a_{9}\).

\[ \begin{aligned} a_{0} + a_{1} + \cdots + a_{9} &= (2 - 1)^{9} = 1 \\ a_{0} &= 2^{9} = 512 \\ a_{9} &= (-1)^{9} = -1 \end{aligned} \]

\[ \begin{aligned} a_{1} + \cdots + a_{8} &= 1 - 512 - (-1) \\ &= -510 \end{aligned} \]

The option \(-511\) forgets to remove \(a_{9}\), and \(-512\) subtracts \(a_{9}\) with the wrong sign.

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