Question

The absolute difference of the coefficient of x7 and x9 in the expansion of $(\frac{2x+1}{2x})^{11}$ is?

• A. 11 x 25
• B. 11 x 27
• C. 11 x 24
• D. 11 x 23

Answer 1

Answer: (B)

11 x 27

Answer 2

The general term of the binomial expansion of $(2x + \frac{1}{2x})^{11}$ is given by:
T(r+1) = (11Cr) (2x)11-r $(\frac{1}{2x})$r
We need to find the absolute difference between the coefficients of x7 and x9 in this expansion.
The coefficient of x7 is the coefficient of T(6), which is given by:
(11 C6) (2x)5 $(\frac{1}{2x})$6 = 462 x5
The coefficient of x9 is the coefficient of T(8), which is given by:
(11 C8) (2x)3 $(\frac{1}{2x})$8 = 165 x3
Therefore, the absolute difference between the coefficients of x7 and x9 is:
|462 x5 - 165 x3| = |297 x3| = 297 |x3|
So, the answer is 11 x 27, as the absolute difference between the coefficients of x7 and x9 is 297, and the power of x is 3.
**Answer. **B