In the expansion of (9x - 1/(3sqrt(x)))^18, x>0, if the term independent of x is (221)k, then k is equal to:

The correct answer is Option 1. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q9 – JEE-MAIN 2026

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Home Maths Binomial TheoremQuestion #7

In the expansion of

{(9 x - \frac{1}{3 \sqrt{x}})}^{18}

x > 0

, if the term independent of

x

(221) k

, then

k

is equal to:

Clue 1 of 3: Binomial Expansion

Recall the general term formula for a binomial expansion $(a + b)^{n}$ .

🧠 Full Solution Path

● Video Walkthrough

Step 1: Identify General Term and Exponent of x✦ Active

The given expression is ${(9 x - \frac{1}{3 \sqrt{x}})}^{18}$ . We can write this as ${(9 x - \frac{1}{3} x^{- 1 / 2})}^{18}$ . Comparing this to $(a + b)^{n}$ , we have $a = 9 x$ , $b = - \frac{1}{3} x^{- 1 / 2}$ , and $n = 18$ . The general term $T_{r + 1}$ is given by:

T_{r + 1} = (\binom{18}{r}) (9 x)^{18 - r} {(- \frac{1}{3} x^{- 1 / 2})}^{r}

Simplify the term to find the exponent of $x$ :

T_{r + 1} = (\binom{18}{r}) 9^{18 - r} x^{18 - r} {(- \frac{1}{3})}^{r} x^{- r / 2}

Combine the powers of $x$ :

T_{r + 1} = (\binom{18}{r}) 9^{18 - r} {(- \frac{1}{3})}^{r} x^{18 - r - r / 2}

Step 2: Find the Value of r for the Term Independent of x○ Expand

For the term independent of $x$ , the exponent of $x$ must be zero. Set the exponent equal to 0 and solve for $r$ :

18 - r - \frac{r}{2} = 0

Multiply by 2 to clear the fraction:

36 - 2 r - r = 0

36 - 3 r = 0

3 r = 36 ⟹ r = 12

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when solving for $r$ .

Step 3: Calculate the Term Independent of x and Solve for k○ Expand

Substitute $r = 12$ back into the coefficient part of the general term:

T_{13} = (\binom{18}{12}) 9^{18 - 12} {(- \frac{1}{3})}^{12}

T_{13} = (\binom{18}{12}) 9^{6} (\frac{1}{3^{12}})

Since $9 = 3^{2}$ , we have $9^{6} = (3^{2})^{6} = 3^{12}$ :

T_{13} = (\binom{18}{12}) 3^{12} (\frac{1}{3^{12}}) = (\binom{18}{12})

Calculate the binomial coefficient $(\binom{18}{12}) = (\binom{18}{6})$ :

(\binom{18}{6}) = \frac{18 \times 17 \times 16 \times 15 \times 14 \times 13}{6 \times 5 \times 4 \times 3 \times 2 \times 1}

Simplifying the expression:

(\binom{18}{6}) = (3 \times 17 \times 4 \times 3 \times 7 \times 13) (after cancelling terms)

(\binom{18}{6}) = 18564

The term independent of $x$ is given as $(221) k$ . So, we have:

18564 = (221) k

Solve for $k$ :

k = \frac{18564}{221} = 84

💡 Teacher's Secret Hint

Remember that $(\binom{n}{r}) = (\binom{n}{n - r})$ can simplify calculations. Double-check your arithmetic for the binomial coefficient.

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Maths Question 9 – JEE-MAIN 2026

Clue 1 of 3: Binomial Expansion

🧠 Full Solution Path