Question

If the coefficient of x in the expansion of \((ax^2+bx+c)(1-2x)^{26\) is \(-56\) and the coefficients of \(x^2\) and \(x^3\) are both zero, then \(a+b+c\) is equal to:}

• A. 1300
• B. 1500
• C. 1403
• D. 1483

Hint:

In problems involving finding coefficients in a product of polynomials, it's not necessary to expand the entire expression.
Focus only on the terms that contribute to the desired powers of x. This saves a lot of time and reduces calculation errors.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a product of a quadratic polynomial \((ax^2+bx+c)\) and a binomial expansion \((1-2x)^{26}\).
We are provided with the coefficients of \(x\), \(x^2\), and \(x^3\) in the resulting expansion.
Our goal is to find the value of the sum of the coefficients of the quadratic, \(a+b+c\).
Step 2: Key Formula or Approach:
We will use the binomial expansion formula for \((1+y)^n\):
\[ (1+y)^n = \binom{n}{0} + \binom{n}{1}y + \binom{n}{2}y^2 + \binom{n}{3}y^3 + \dots \] In our case, \(y = -2x\) and \(n=26\).
\[ (1-2x)^{26} = \binom{26}{0} + \binom{26}{1}(-2x) + \binom{26}{2}(-2x)^2 + \binom{26}{3}(-2x)^3 + \dots \] \[ (1-2x)^{26} = 1 - 2\binom{26}{1}x + 4\binom{26}{2}x^2 - 8\binom{26}{3}x^3 + \dots \] Step 3: Detailed Explanation:
Let's calculate the first few terms of the expansion of \((1-2x)^{26}\):
Coefficient of \(x^0\) is \(\binom{26}{0} = 1\).
Coefficient of \(x^1\) is \(\binom{26}{1}(-2) = 26 \times (-2) = -52\).
Coefficient of \(x^2\) is \(\binom{26}{2}(-2)^2 = \frac{26 \times 25}{2} \times 4 = 13 \times 25 \times 4 = 1300\).
Coefficient of \(x^3\) is \(\binom{26}{3}(-2)^3 = \frac{26 \times 25 \times 24}{3 \times 2 \times 1} \times (-8) = 26 \times 25 \times 4 \times (-8) = -20800\).
So, \((1-2x)^{26} = 1 - 52x + 1300x^2 - 20800x^3 + \dots\).
Now we multiply this by \((ax^2+bx+c)\):
\[ (ax^2+bx+c)(1 - 52x + 1300x^2 - 20800x^3 + \dots) \] Let's find the coefficients in the product:
Coefficient of x: This is obtained by \((c \times \text{coeff of } x) + (b \times \text{coeff of } x^0)\).
\[ c(-52) + b(1) = -56 \implies b - 52c = -56 \quad \text{(Eq. 1)} \] Coefficient of \(x^2\): This is obtained by \((c \times \text{coeff of } x^2) + (b \times \text{coeff of } x) + (a \times \text{coeff of } x^0)\).
\[ c(1300) + b(-52) + a(1) = 0 \implies a - 52b + 1300c = 0 \quad \text{(Eq. 2)} \] Coefficient of \(x^3\): This is obtained by \((c \times \text{coeff of } x^3) + (b \times \text{coeff of } x^2) + (a \times \text{coeff of } x)\).
\[ c(-20800) + b(1300) + a(-52) = 0 \] Dividing by -52, we get:
\[ a - 25b + 400c = 0 \quad \text{(Eq. 3)} \] Now we solve the system of equations. Subtract Eq. 3 from Eq. 2:
\[ (a - 52b + 1300c) - (a - 25b + 400c) = 0 - 0 \] \[ -27b + 900c = 0 \implies 27b = 900c \implies b = \frac{900}{27}c = \frac{100}{3}c \] Substitute \(b\) in Eq. 1:
\[ \frac{100}{3}c - 52c = -56 \] \[ (\frac{100 - 156}{3})c = -56 \] \[ \frac{-56}{3}c = -56 \implies c = 3 \] Now find \(b\):
\[ b = \frac{100}{3} \times 3 = 100 \] Now find \(a\) using Eq. 3:
\[ a - 25(100) + 400(3) = 0 \] \[ a - 2500 + 1200 = 0 \] \[ a - 1300 = 0 \implies a = 1300 \] We need to find \(a+b+c\):
\[ a+b+c = 1300 + 100 + 3 = 1403 \] Step 4: Final Answer:
The values are \(a=1300\), \(b=100\), and \(c=3\). The sum \(a+b+c\) is \(1403\).