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Mathematics · Binomial Theorem

JEE Main 2026 — 21 January, Morning Shift — Question 16

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

  1. Option A:

    1300

  2. Option B:

    1500

  3. Option C:

    1403

    Correct
  4. Option D:

    1483

Answer: C

Step-by-step solution

(ax2+bc+c)∑r=02626Cr(−2x)r\left(a x^{2}+b c+c\right) \sum_{\mathrm{r}=0}^{26}{ }^{26} \mathrm{C}_{\mathrm{r}}(-2 \mathrm{x})^{\mathrm{r}}

Coeff. of x2:a.26C0(−2)0+b.26C1(−2)+\mathrm{x}^{2}: \mathrm{a} .{ }^{26} \mathrm{C}_{0}(-2)^{0}+\mathrm{b} .{ }^{26} \mathrm{C}_{1}(-2)+ c. 26C2(−2)2=0{ }^{26} \mathrm{C}_{2}(-2)^{2}=0 ⇒a−52b+1300c=0\begin{gathered} \Rightarrow a-52 b+1300 c=0 \end{gathered} Coeff. of x3:a.26C1(−2)+b.26C2(−2)2+x^{3}: a .{ }^{26} C_{1}(-2)+b .{ }^{26} C_{2}(-2)^{2}+ c. 26C3(−2)3=0{ }^{26} \mathrm{C}_{3}(-2)^{3}=0 ⇒−52a+1300b−20800c=0\begin{gathered} \Rightarrow-52 a+1300 b-20800 c=0 \end{gathered} Coeff. of x=−56x=-56 ⇒b.26C0(−2)0+c.26C1(−2)1=−56\Rightarrow \mathrm{b} .{ }^{26} \mathrm{C}_{0}(-2)^{0}+\mathrm{c} .{ }^{26} \mathrm{C}_{1}(-2)^{1}=-56 b−52c=−56\begin{gathered} b-52 c=-56 \end{gathered}

After solving (1), (2) & (3)

a=1300, b=100,c=3\mathrm{a}=1300, \mathrm{~b}=100, \mathrm{c}=3

⇒a+b+c=1403\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1403

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
21 January, Morning Shift
Subject
Mathematics
Chapter
Binomial Theorem
Topic
Binomial Coefficients