Binomial TheoremeasyPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Term Independent of x gives k = 84 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
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
A8484correct
B7878
C168168
D198198
Solution
Step 1: General term:
Tr+1=18Cr(9x)18−r(−13x)r=(−13)r18Cr 918−r x18−3r2.T_{r+1}={}^{18}C_r(9x)^{18-r}\left(-\frac{1}{3\sqrt{x}}\right)^r=\left(-\frac13\right)^r{}^{18}C_r\,9^{18-r}\,x^{18-\frac{3r}{2}}.
 Step 2: For the term independent of xx: 18−3r2=0⇒r=1218-\dfrac{3r}{2}=0\Rightarrow r=12. Step 3: Its coefficient:
(−13)1218C12 918−12=1312⋅18C12⋅312=18C12=18564.\left(-\frac13\right)^{12}{}^{18}C_{12}\,9^{18-12}=\frac{1}{3^{12}}\cdot{}^{18}C_{12}\cdot 3^{12}={}^{18}C_{12}=18564.
 Step 4: Given this equals 221k221k:
18564=221k ⇒ k=18564221=84.18564=221k\ \Rightarrow\ k=\frac{18564}{221}=84.
 Correct answer: (1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Binomial Theorem · medium
The coefficient of x2x^2 in the expansion of (2x2+1x)10\left(2x^2+\dfrac{1}{x}\right)^{10}, x≠0x\neq0, is:
Binomial Theorem · medium
Let t1=7t_1=7, t2=77t_2=7^7, t3=777t_3=7^{7^7}, and so on. The digit at the tens place of t1000t_{1000} is
Binomial Theorem · easy
If the sum of the coefficients of x7x^7 and x14x^{14} in the expansion of (1x3−x4)n\left(\dfrac{1}{x^3}-x^4\right)^n, x≠0x\ne0, is zero, then the value of nn is
Binomial Theorem · medium
Let the smallest value of k∈Nk\in\mathbb{N}, for which the coefficient of x3x^3 in (1+x)3+(1+x)4+(1+x)5+⋯+(1+x)99+(1+kx)100(1+x)^3+(1+x)^4+(1+x)^5+\cdots+(1+x)^{99}+(1+kx)^{100}, x≠0x\ne0, is (43n+1014)(100C3)\left(43n+\dfrac{101}{4}\right)\big({}^{100}C_3\big) for some n∈Nn\in\mathbb{N}, be pp. Then the value of p+np+n is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath