If the coefficient of x in the expansion of (ax² + bx + c)(1 - 2x)²⁶$(ax^{2} + bx + c)(1 - 2x)^{26}$ is -56 and the coefficients of x²$x^{2}$ and x³$x^{3}$ are both zero, then a + b + c$a + b + c$ is equal to

Related Formula

The coefficient extraction utilizes the distributive property over the polynomial (ax²+bx+c)$(ax^2+bx+c)$.

Core Logic

Expansion of expression: (ax² + bx + c) Σr=0²⁶ 26r (-2x)r$(ax^{2} + bx + c) \sum_{r=0}^{26} \binom{26}{r} (-2x)^{r}$

For the term in x²$x^2$, we pick contributions yielding total power 2: a · (x⁰ term of sum) + b · (x¹ term) + c · (x² term) = 0$a \cdot (x^0 \text{ term of sum}) + b \cdot (x^1 \text{ term}) + c \cdot (x^2 \text{ term}) = 0$ a · 260(-2)⁰ + b · 261(-2)¹ + c · 262(-2)² = 0$a \cdot \binom{26}{0}(-2)^0 + b \cdot \binom{26}{1}(-2)^1 + c \cdot \binom{26}{2}(-2)^2 = 0$ a - 52b + 1300c = 0 (1)$a - 52b + 1300c = 0 \dots(1)$

Step 1: Formulate the system of linear equations

For the term in x³$x^3$ (set to 0): a · (x¹ term) + b · (x² term) + c · (x³ term) = 0$a \cdot (x^1 \text{ term}) + b \cdot (x^2 \text{ term}) + c \cdot (x^3 \text{ term}) = 0$ a · 261(-2)¹ + b · 262(-2)² + c · 263(-2)³ = 0$a \cdot \binom{26}{1}(-2)^1 + b \cdot \binom{26}{2}(-2)^2 + c \cdot \binom{26}{3}(-2)^3 = 0$ -52a + 1300b - 20800c = 0 (2)$-52a + 1300b - 20800c = 0 \dots(2)$

For the term in x¹$x^1$ (set to -56): b · (x⁰ term) + c · (x¹ term) = -56$b \cdot (x^0 \text{ term}) + c \cdot (x^1 \text{ term}) = -56$ b · 260(-2)⁰ + c · 261(-2)¹ = -56$b \cdot \binom{26}{0}(-2)^0 + c \cdot \binom{26}{1}(-2)^1 = -56$ b - 52c = -56 (3)$b - 52c = -56 \dots(3)$

Step 2: Solve the Linear System

From (3), b = 52c - 56$b = 52c - 56$. Substitute into (2) divided by -52$-52$ to simplify: a - 25b + 400c = 0$a - 25b + 400c = 0$. Substitute into (1): a - 52b + 1300c = 0$a - 52b + 1300c = 0$. Subtracting: 27b - 900c = 0 ⇒ b = (100)/(3)c$27b - 900c = 0 \Rightarrow b = \frac{100}{3}c$. Wait, recalculating directly: From (2): -52a + 1300b - 20800c = 0 ⇒ a - 25b + 400c = 0$-52a + 1300b - 20800c = 0 \Rightarrow a - 25b + 400c = 0$. From (1): a - 52b + 1300c = 0$a - 52b + 1300c = 0$. (a - 25b + 400c) - (a - 52b + 1300c) = 0 ⇒ 27b - 900c = 0 ⇒ 3b = 100c$ (a - 25b + 400c) - (a - 52b + 1300c) = 0 \Rightarrow 27b - 900c = 0 \Rightarrow 3b = 100c$. Substitute into (3): b - 52(3b/100) = -56 ⇒$b - 52(3b/100) = -56 \Rightarrow$ fractional results? Let's check coefficients accurately. 263(-2)³ = (26 · 25 · 24)/(6) × (-8) = 2600 × -8 = -20800$\binom{26}{3}(-2)^3 = \frac{26 \cdot 25 \cdot 24}{6} \times (-8) = 2600 \times -8 = -20800$. Divide by -52: a - 25b + 400c = 0$a - 25b + 400c = 0$. Correct. (a - 52b + 1300c) - (a - 25b + 400c) = -27b + 900c = 0 ⇒ 27b = 900c ⇒ b = (100)/(3)c$(a - 52b + 1300c) - (a - 25b + 400c) = -27b + 900c = 0 \Rightarrow 27b = 900c \Rightarrow b = \frac{100}{3}c$. This contradicts integer expectations. Let's substitute b$b$ from (3) directly. b = 52c - 56$b = 52c - 56$. 27(52c - 56) = 900c$27(52c - 56) = 900c$ 1404c - 1512 = 900c ⇒ 504c = 1512 ⇒ c = 3$1404c - 1512 = 900c \Rightarrow 504c = 1512 \Rightarrow c = 3$. Then b = 52(3) - 56 = 156 - 56 = 100$b = 52(3) - 56 = 156 - 56 = 100$. From (1): a - 52(100) + 1300(3) = 0 ⇒ a - 5200 + 3900 = 0 ⇒ a = 1300$a - 52(100) + 1300(3) = 0 \Rightarrow a - 5200 + 3900 = 0 \Rightarrow a = 1300$.

Step 3: Compute final value

We have a = 1300$a = 1300$, b = 100$b = 100$, c = 3$c = 3$.

Pattern Recognition

Polynomial-Binomial product coefficient extractions strictly generate cascaded linear Diophantine-style equations. Solving from the lowest degree constraint (x¹$x^1$) upward sequentially unwinds the system cleanly.

Chapter Mix

Class 11 Maths: Binomial Theorem