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JEE Main 20264 AprilEvening Shift
Exam
JEE Main
Year
2026
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Binomial Theorem

JEE Main 2026 · 4 April · Evening Shift · Mathematics · Binomial Theorem

In the expansion of (9x- 1 3 x )^ 18 , x>0 , if the term...

In the expansion of (9x−13x)18\left(9x-\dfrac{1}{3\sqrt{x}}\right)^{18}, x>0x>0, if the term independent of xx is (221)k(221)k, then kk is equal to:

Options

  1. A8484
  2. B7878
  3. C168168
  4. D198198

Correct answer

A. 8484

Step-by-step solution

The general term in the expansion of (9x−13x)18\left(9x - \dfrac{1}{3\sqrt{x}}\right)^{18} is given by:

Tr+1=18Cr(9x)18−r(−13x)rT_{r+1} = ^{18}C_{r} (9x)^{18-r} \left(-\dfrac{1}{3\sqrt{x}}\right)^r

Tr+1=18Cr918−r(−13)rx18−rx−r/2T_{r+1} = ^{18}C_{r} 9^{18-r} \left(-\dfrac{1}{3}\right)^r x^{18-r} x^{-r/2}

Tr+1=18Cr918−r(−13)rx18−3r2T_{r+1} = ^{18}C_{r} 9^{18-r} \left(-\dfrac{1}{3}\right)^r x^{18 - \dfrac{3r}{2}}

For the term independent of xx, the exponent of xx must be zero:

18−3r2=018 - \dfrac{3r}{2} = 0

3r2=18⇒r=12\dfrac{3r}{2} = 18 \Rightarrow r = 12

Substituting r=12r = 12 into the general term:

T13=18C12918−12(−13)12T_{13} = ^{18}C_{12} 9^{18-12} \left(-\dfrac{1}{3}\right)^{12}

T13=18C1296(13)12T_{13} = ^{18}C_{12} 9^6 \left(\dfrac{1}{3}\right)^{12}

T13=18C12(32)61312T_{13} = ^{18}C_{12} (3^2)^6 \dfrac{1}{3^{12}}

T13=18C123121312=18C12T_{13} = ^{18}C_{12} 3^{12} \dfrac{1}{3^{12}} = ^{18}C_{12}

Now, calculating 18C12^{18}C_{12}:

18C12=18C6=18×17×16×15×14×136×5×4×3×2×1^{18}C_{12} = ^{18}C_{6} = \dfrac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1}

18C12=17×13×3×2×14^{18}C_{12} = 17 \times 13 \times 3 \times 2 \times 14

18C12=221×84^{18}C_{12} = 221 \times 84

Given that the term independent of xx is 221k221k:

221k=221×84221k = 221 \times 84

k=84k = 84

Answer: 8484
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