Find the general term in the expansion of (2x - 3/x)⁹

medium CBSE JEE-MAIN NCERT Class 11 3 min read

Question

Find the general term in the expansion of (2x3x)9\left(2x - \dfrac{3}{x}\right)^9.

(NCERT Class 11, Exercise 8.2)


Solution — Step by Step

For (a+b)n(a + b)^n, the general term is:

Tr+1=(nr)anrbrT_{r+1} = \binom{n}{r} a^{n-r} b^r

Here a=2xa = 2x, b=3xb = -\dfrac{3}{x}, and n=9n = 9.

Tr+1=(9r)(2x)9r(3x)rT_{r+1} = \binom{9}{r} (2x)^{9-r} \left(-\frac{3}{x}\right)^r =(9r)29rx9r(3)rxr= \binom{9}{r} \cdot 2^{9-r} \cdot x^{9-r} \cdot (-3)^r \cdot x^{-r} Tr+1=(9r)29r(3)rx92r\boxed{T_{r+1} = \binom{9}{r} \cdot 2^{9-r} \cdot (-3)^r \cdot x^{9-2r}}

where r=0,1,2,,9r = 0, 1, 2, \ldots, 9.

The power of xx is 92r9 - 2r. This tells us everything: the term independent of xx occurs when 92r=09 - 2r = 0, i.e., r=4.5r = 4.5 — which is not an integer, so there is no constant term in this expansion.


Why This Works

The binomial theorem systematically distributes the choice: in each of the 9 factors of (2x3/x)(2x - 3/x), we pick either 2x2x or 3/x-3/x. The (9r)\binom{9}{r} counts the number of ways to pick 3/x-3/x from exactly rr of the 9 factors (and 2x2x from the remaining 9r9-r).

The power of xx in each term is (9r)r=92r(9-r) - r = 9 - 2r because each 2x2x contributes x1x^1 and each 3/x-3/x contributes x1x^{-1}. This decreasing power pattern is useful for finding specific terms — the term with x3x^3, the constant term, etc.


Alternative Method — Find a specific term

If the question asks for the coefficient of x3x^3: set 92r=39 - 2r = 3, so r=3r = 3.

T4=(93)26(3)3x3=84×64×(27)x3=145152x3T_4 = \binom{9}{3} \cdot 2^6 \cdot (-3)^3 \cdot x^3 = 84 \times 64 \times (-27) \cdot x^3 = -145152\,x^3

In JEE, binomial questions often ask for a specific term — “find the coefficient of x5x^5” or “find the term independent of xx.” Set the exponent of xx equal to the desired power and solve for rr. If rr isn’t a non-negative integer n\leq n, that term doesn’t exist.


Common Mistake

Students frequently drop the negative sign from (3/x)r(-3/x)^r. The term b=3/xb = -3/x, not 3/x3/x. When rr is odd, (3)r(-3)^r is negative, and when rr is even, it’s positive. Forgetting this sign alternation will give wrong coefficients for odd-rr terms. Always write (3)r(-3)^r explicitly and simplify the sign at the end.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next