Question
Find the coefficient of x7 in the expansion of (2x−3/x)11.
Solution — Step by Step
In (a+b)n, the general term is Tr+1=(rn)an−rbr.
Here a=2x, b=−3/x, n=11:
Tr+1=(r11)(2x)11−r(−3/x)r.
(2x)11−r=211−rx11−r.
(−3/x)r=(−3)rx−r.
So Tr+1=(r11)211−r(−3)rx11−r−r=(r11)211−r(−3)rx11−2r.
11−2r=7⟹r=2.
T3=(211)⋅29⋅(−3)2⋅x7.
(211)=55. 29=512. (−3)2=9.
Coefficient = 55×512×9=55×4608=253,440.
Final answer: Coefficient of x7 is 253,440.
Why This Works
The general term technique handles any “find the coefficient of xk” problem in a binomial expansion. The strategy is always:
- Write Tr+1 in terms of r.
- Set the exponent of x equal to the target.
- Solve for r (must be a non-negative integer ≤n).
- Plug back to get the coefficient.
If r comes out non-integer, the term doesn’t exist — coefficient is zero.
Alternative Method
We could expand a few terms manually, but with n=11, this is impractical. The general-term shortcut is the only reasonable approach.
Common Mistake
The classic slip: forgetting the negative sign in b=−3/x. Students write b=3/x and miss the (−1)r factor. Here r=2 is even, so the sign happens to be positive — but for r odd (e.g., coefficient of x9 would have r=1), the sign would matter and you’d get a wrong answer.
Also, watch the exponent algebra: x11−r/xr=x11−2r, not x11−r−r written wrong as x11−r/r.