Question

If \( (2 - x)^9 = a_0 + a_1x + \dots + a_9x^9 \), find \( a_1 + a_2 + \dots + a_8 \)

Hint:

To find the sum of coefficients in a polynomial, substitute \(x = 1\) into the polynomial and simplify.

Answer 1

Step 1: Expand the binomial using the binomial theorem.
The given equation is \( (2 - x)^9 \). By using the binomial expansion, we expand it as: \[ (2 - x)^9 = \sum_{k=0}^{9} \binom{9}{k} 2^{9-k} (-x)^k \] This gives: \[ = \sum_{k=0}^{9} \binom{9}{k} 2^{9-k} (-1)^k x^k \]
Step 2: Find the sum \(a_1 + a_2 + \dots + a_8\).
The coefficients \(a_1, a_2, \dots, a_8\) correspond to the terms for \(x^1, x^2, \dots, x^8\) in the expansion. To find \( a_1 + a_2 + \dots + a_8 \), we need to evaluate the sum of the coefficients of the terms from \(x^1\) to \(x^8\). By substituting \(x = 1\) into the expansion, we get: \[ (2 - 1)^9 = a_0 + a_1 + \dots + a_9 \] \[ 1^9 = a_0 + a_1 + \dots + a_9 \] So: \[ 1 = a_0 + a_1 + \dots + a_9 \]
Step 3: Subtract \(a_9\) from both sides.
To find \( a_1 + a_2 + \dots + a_8 \), subtract \(a_9\) from both sides: \[ a_1 + a_2 + \dots + a_8 = 1 - a_9 \] We can calculate \(a_9\) using the binomial expansion, where \(k = 9\): \[ a_9 = \binom{9}{9} 2^0 (-1)^9 = -1 \] Thus: \[ a_1 + a_2 + \dots + a_8 = 1 - (-1) = 1 + 1 = 2 \] Thus, the sum of the coefficients is: \[ \boxed{2} \]