Binomial Theorem – Coefficients – JEE Main 21 Jan 2026 Shift 1

Solution:

Let the expansion be $E = (ax^2 + bx + c) \sum_{r=0}^{26} {}^{26}C_r (-2x)^r$.
We collect the coefficients of $x, x^2, x^3$ and equate them to given values.

**Coefficient of $x^2$ is 0:**
The term involving $x^2$ is formed by:
$a(1) \cdot (\text{const in expansion}) + b(x) \cdot (\text{coeff of } x) + c(x^2) \cdot (\text{coeff of } x^2)$
$$a \cdot {}^{26}C_0 + b \cdot {}^{26}C_1(-2) + c \cdot {}^{26}C_2(-2)^2 = 0$$
$$a(1) + b(26)(-2) + c \left(\frac{26 \times 25}{2}\right)(4) = 0$$
$$a – 52b + 1300c = 0 \quad \dots(1)$$

**Coefficient of $x^3$ is 0:**
The term involving $x^3$ is formed by:
$a(1) \cdot (\text{coeff of } x) + b(x) \cdot (\text{coeff of } x^2) + c(x^2) \cdot (\text{coeff of } x^3)$
$$a \cdot {}^{26}C_1(-2) + b \cdot {}^{26}C_2(-2)^2 + c \cdot {}^{26}C_3(-2)^3 = 0$$
$$a(-52) + b(1300) + c \left(\frac{26 \times 25 \times 24}{6}\right)(-8) = 0$$
$$-52a + 1300b – 20800c = 0$$
Divide by 52:
$$-a + 25b – 400c = 0 \quad \dots(2)$$

**Coefficient of $x$ is -56:**
$$b \cdot {}^{26}C_0 + c \cdot {}^{26}C_1(-2) = -56$$
$$b(1) – 52c = -56 \quad \dots(3)$$

**Solving the equations:**
Add (1) and (2):
$$(a – 52b + 1300c) + (-a + 25b – 400c) = 0$$
$$-27b + 900c = 0 \Rightarrow 27b = 900c \Rightarrow 3b = 100c \Rightarrow b = \frac{100}{3}c$$

Substitute $b = \frac{100}{3}c$ into (3):
$$\frac{100}{3}c – 52c = -56$$
$$\frac{100c – 156c}{3} = -56$$
$$-\frac{56c}{3} = -56 \Rightarrow c = 3$$

Find $b$:
$$b = \frac{100}{3}(3) = 100$$

Find $a$ using (1):
$$a – 52(100) + 1300(3) = 0$$
$$a – 5200 + 3900 = 0$$
$$a – 1300 = 0 \Rightarrow a = 1300$$

Calculate $a+b+c$:
$$a+b+c = 1300 + 100 + 3 = 1403$$

Ans. (3)