Binomial Theorem Part 3

Concept:We use binomial expansion of $(1-2x)^{26}$ and multiply by $(ax^2+bx+c)$.Coefficients of $x$, $x^2$, and $x^3$ in the product are equated to given values.Explanation:The product is: $a x^2(1-2x)^{26} + b x(1-2x)^{26} + c(1-2x)^{26}$.General term of $(1-2x)^{26}$ is $\binom{26}{r}(-2x)^r = \binom{26}{r}(-2)^r x^r$.Coefficient of $x$:Only $bx(1-2x)^{26}$ and $c(1-2x)^{26}$ contribute.From $bx$ term: $b \cdot \binom{26}{0} = b$.From $c$ term: $c \cdot \binom{26}{1}(-2) = -52c$.So coefficient: $b - 52c = -56$ (given).Coefficient of $x^2$:Contributions: $a x^2$ term: $a \cdot \binom{26}{0} = a$.$bx$ term: $b \cdot \binom{26}{1}(-2) = -52b$.$c$ term: $c \cdot \binom{26}{2}(-2)^2 = c \cdot 325 \cdot 4 = 1300c$.So coefficient: $a - 52b + 1300c = 0$ (given).Coefficient of $x^3$:Contributions: $a x^2$ term: $a \cdot \binom{26}{1}(-2) = -52a$.$bx$ term: $b \cdot \binom{26}{2}(-2)^2 = b \cdot 325 \cdot 4 = 1300b$.$c$ term: $c \cdot \binom{26}{3}(-2)^3 = c \cdot 2600 \cdot (-8) = -20800c$.So coefficient: $-52a + 1300b - 20800c = 0$ (given).The three equations are:1. $b - 52c = -56$2. $a - 52b + 1300c = 0$3. $-52a + 1300b - 20800c = 0$Simplify equation 3 by dividing by 52: $-a + 25b - 400c = 0$.Now solve: from eq1, $b = 52c -56$. Substitute into eq2 and eq3.From eq2: $a - 52(52c-56) + 1300c = 0 \Rightarrow a - 2704c + 2912 + 1300c = 0 \Rightarrow a - 1404c + 2912 = 0$ → (A)From eq3: $-a + 25(52c-56) - 400c = 0 \Rightarrow -a + 1300c - 1400 - 400c = 0 \Rightarrow -a + 900c - 1400 = 0 \Rightarrow a = 900c - 1400$ → (B)Equate (A) and (B): $900c - 1400 = 1404c - 2912$$\Rightarrow -1400 + 2912 = 1404c - 900c \Rightarrow 1512 = 504c \Rightarrow c = 3$.Then $b = 52(3)-56 = 156-56 = 100$.$a = 900(3)-1400 = 2700-1400 = 1300$.Thus $a+b+c = 1300 + 100 + 3 = 1403$.Answer:$1403$ (Option D)