Question:
The absolute difference of the coefficient of x7 and x9 in the expansion of \((\frac{2x+1}{2x})^{11}\) is?
The absolute difference of the coefficient of x7 and x9 in the expansion of \((\frac{2x+1}{2x})^{11}\) is?
Updated On: Sep 24, 2024
- 11 x 25
- 11 x 27
- 11 x 24
- 11 x 23
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The Correct Option is B
Solution and Explanation
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
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
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Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .