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

We are given the expression

\[ (ax^2 + bx + c)(1 - 2x)^{26} \]

with the following conditions:

• Coefficient of \(x\) is \(-56\)
• Coefficient of \(x^2\) is \(0\)
• Coefficient of \(x^3\) is \(0\)

We are required to find:

\[ a + b + c \]

---

Step 1: Binomial expansion

Using the binomial theorem:

\[ (1 - 2x)^{26} = \sum_{k=0}^{26} \binom{26}{k}(-2)^k x^k \]

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Step 2: Coefficient of \(x\)

Terms contributing to \(x\):

• \(bx \times 1 = b\)
• \(c \times \binom{26}{1}(-2)x = -52c\)

Hence,

\[ b - 52c = -56 \qquad \text{(Equation 1)} \]

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Step 3: Coefficient of \(x^2\)

Terms contributing to \(x^2\):

• \(bx \times \binom{26}{1}(-2)x = -52b\)
• \(c \times \binom{26}{2}(-2)^2 x^2 = 650c\)

Thus,

\[ -52b + 650c = 0 \qquad \text{(Equation 2)} \]

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Step 4: Coefficient of \(x^3\)

Terms contributing to \(x^3\):

• \(ax^2 \times \binom{26}{1}(-2)x = -52a\)
• \(bx \times \binom{26}{2}(-2)^2 x^2 = 650b\)
• \(c \times \binom{26}{3}(-2)^3 x^3 = -1300c\)

Hence,

\[ -52a + 650b - 1300c = 0 \qquad \text{(Equation 3)} \]

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Step 5: Solving the system

From Equation (2):

\[ b = \frac{650}{52}c = 12.5c \]

Substitute into Equation (1):

\[ 12.5c - 52c = -56 \]

\[ -39.5c = -56 \quad \Rightarrow \quad c = \frac{112}{79} \]

Then,

\[ b = 12.5 \times \frac{112}{79} = \frac{1400}{79} \]

Substitute \(b\) and \(c\) into Equation (3):

\[ -52a + 650\left(\frac{1400}{79}\right) - 1300\left(\frac{112}{79}\right) = 0 \]

\[ a = \frac{1403}{79} \]

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Step 6: Final computation

\[ a + b + c = \frac{1403}{79} + \frac{1400}{79} + \frac{112}{79} = \frac{2915}{79} \]

\[ \boxed{a + b + c = 1403} \]