Question
Find the term independent of x in the expansion of (2x2−x1)12.
Solution — Step by Step
The (r+1)-th term in the expansion of (a+b)n is Tr+1=(rn)an−rbr. Here a=2x2, b=−1/x, n=12:
Tr+1=(r12)(2x2)12−r(−x1)r
Tr+1=(r12)212−r(−1)rx2(12−r)x−r=(r12)212−r(−1)rx24−3r
For the term to be independent of x:
24−3r=0⟹r=8
T9=(812)24(−1)8=495×16×1=7920
The term independent of x is 7920.
Why This Works
The general-term formula lets us isolate the exponent of x as a function of r. Setting that exponent to zero gives us the index of the term we want, and substituting back gives its coefficient.
This technique generalises: to find the coefficient of xk, set the exponent expression equal to k and solve for r.
Alternative Method
Multinomial reasoning: the term comes from picking b=−1/x exactly r times and a=2x2 exactly 12−r times. The combined exponent of x is 2(12−r)−r=24−3r, set to zero. Same logic, no formula recall needed.
Common Mistake
Forgetting the sign (−1)r from the negative term in b. Here r=8 is even, so (−1)8=+1, but for r=9,11 etc., the sign matters. JEE Main 2023 had a question where the trap option used the wrong sign.