JEE Main 2026 — Mathematics Algebra

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

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

$$T_{r+1} = ^{18}C_{r} 9^{18-r} \left(-\dfrac{1}{3}\right)^r x^{18-r} x^{-r/2}$$Tr+1​=18Cr​918−r(−31​)rx18−rx−r/2

$$T_{r+1} = ^{18}C_{r} 9^{18-r} \left(-\dfrac{1}{3}\right)^r x^{18 - \dfrac{3r}{2}}$$Tr+1​=18Cr​918−r(−31​)rx18−23r​

For the term independent of $x$x, the exponent of $x$x must be zero:

$$18 - \dfrac{3r}{2} = 0$$18−23r​=0

$$\dfrac{3r}{2} = 18 \Rightarrow r = 12$$23r​=18⇒r=12

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

$$T_{13} = ^{18}C_{12} 9^{18-12} \left(-\dfrac{1}{3}\right)^{12}$$T13​=18C12​918−12(−31​)12

$$T_{13} = ^{18}C_{12} 9^6 \left(\dfrac{1}{3}\right)^{12}$$T13​=18C12​96(31​)12

$$T_{13} = ^{18}C_{12} (3^2)^6 \dfrac{1}{3^{12}}$$T13​=18C12​(32)63121​

$$T_{13} = ^{18}C_{12} 3^{12} \dfrac{1}{3^{12}} = ^{18}C_{12}$$T13​=18C12​3123121​=18C12​

Now, calculating $^{18}C_{12}$18C12​:

$$^{18}C_{12} = ^{18}C_{6} = \dfrac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1}$$18C12​=18C6​=6×5×4×3×2×118×17×16×15×14×13​

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

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

Given that the term independent of $x$x is $221k$221k:

$$221k = 221 \times 84$$221k=221×84

$$k = 84$$k=84

Answer: $84$84