Question

Let $(2 - x)^9 = a_0 + a_1x + a_2x^2 + \dots + a_9x^9$. Then the value of $a_1 + a_2 + a_3 + \dots + a_8$ is equal to

• A. -511
• B. 510
• C. -512
• D. 512
• E. -510

Hint:

Sum of coefficients excluding extreme terms is simply \( P(1) - a_0 - a_n \). Quick calculation: \( 1 - 512 - (-1) = -510 \).

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
To find sums of coefficients in a binomial expansion, we substitute specific values for $x$. Putting $x=1$ gives the sum of all coefficients.

Step 2: Detailed Explanation:
1. Sum of all coefficients:
Put $x = 1$ in $(2 - x)^9$:
\[ (2 - 1)^9 = a_0 + a_1 + a_2 + \dots + a_8 + a_9 \]
\[ 1 = a_0 + (a_1 + a_2 + \dots + a_8) + a_9 \]
2. Identify specific coefficients:
- $a_0$ is the constant term (when $x=0$):
\[ a_0 = (2 - 0)^9 = 2^9 = 512 \]
- $a_9$ is the coefficient of $x^9$:
Using general term $T_{r+1} = \binom{9}{r} (2)^{9-r} (-x)^r$:
For $r=9$, $T_{10} = \binom{9}{9} (2)^0 (-1)^9 x^9 = -1x^9$.
So, $a_9 = -1$.
3. Calculate the required sum:
\[ 1 = 512 + (a_1 + a_2 + \dots + a_8) + (-1) \]
\[ 1 = 511 + (a_1 + a_2 + \dots + a_8) \]
\[ a_1 + a_2 + \dots + a_8 = 1 - 511 = -510 \]

Step 3: Final Answer:
The sum is -510.