JEE Main 2026 (Online) 21st January Morning ShiftQuestion 1
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 :
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. Expand $(1-2x)^{26}$ up to $x^3$, then match the coefficients of $x$, $x^2$, $x^3$. Formulas used. (1) $(1+y)^n=1+ny+{}^{n}C_2\,y^2+{}^{n}C_3\,y^3+\cdots$Step 1 - First terms. With $y=-2x$, $n=26$: $$(1-2x)^{26}=1-52x+1300x^2-20800x^3+\cdots$$ since ${}^{26}C_2\cdot4=325\cdot4=1300$ and ${}^{26}C_3\cdot8=2600\cdot8=20800$. Step 2 - Equations. Multiply by $ax^2+bx+c$: $$x:\ b-52c=-56$$ $$x^2:\ a-52b+1300c=0$$ $$x^3:\ -52a+1300b-20800c=0\ \Longrightarrow\ a-25b+400c=0$$ Step 3 - Solve. Subtract the $x^2$ equation from the $x^3$ one: $27b=900c$, so $b=\frac{100c}{3}$. Then $$\frac{100c}{3}-52c=-56\ \Longrightarrow\ -\frac{56c}{3}=-56\ \Longrightarrow\ c=3.$$ So $b=100$ and $a=52b-1300c=5200-3900=1300$. Step 4 - Value. $a+b+c=1300+100+3=1403$. Answer: (D)