The absolute difference of the coefficients of \(x^{10}\) and \(x^7\) in the expansion of \(\left(2x^2 + \frac{1}{2x}\right)^{11}\) is equal to:
When dealing with expansions, simplify the general term first, then identify the required terms using the exponent relationships.
To find the absolute difference of the coefficients of \(x^{10}\) and \(x^7\) in the expansion of \(\left(2x^2 + \frac{1}{2x}\right)^{11}\), we need to focus on the general term in the binomial expansion of \((a+b)^n\), which is given by:
\(T_k = \binom{n}{k} a^{n-k} b^k\)
For our problem, \(a = 2x^2\) and \(b = \frac{1}{2x}\), with \(n = 11\).
Thus, the general term \(T_k\) becomes:
\(T_k = \binom{11}{k} (2x^2)^{11-k} \left(\frac{1}{2x}\right)^k = \binom{11}{k} \cdot 2^{11-k} \cdot x^{2(11-k)} \cdot \frac{1}{2^k} \cdot x^{-k}\)
Simplifying the term further, we have:
\(T_k = \binom{11}{k} \cdot \frac{2^{11-k}}{2^k} \cdot x^{22-2k-k} = \binom{11}{k} \cdot 2^{11-2k} \cdot x^{22-3k}\)
To find the coefficient of \(x^{10}\), set the power of \(x\) to 10:
\(22 - 3k = 10\)
\(3k = 12\)
\(k = 4\)
So, the term is:
\(T_4 = \binom{11}{4} \cdot 2^{11-8} \cdot x^{10} = \binom{11}{4} \cdot 2^3\)
Calculate the coefficient:
\(\binom{11}{4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330\)
\(2^3 = 8\)
The coefficient of \(x^{10}\) is \(330 \cdot 8 = 2640\).
To find the coefficient of \(x^7\), set the power of \(x\) to 7:
\(22 - 3k = 7\)
\(3k = 15\)
\(k = 5\)
So, the term is:
\(T_5 = \binom{11}{5} \cdot 2^{11-10} \cdot x^7 = \binom{11}{5} \cdot 2^1\)
Calculate the coefficient:
\(\binom{11}{5} = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 462\)
The coefficient of \(x^7\) is \(462 \cdot 2 = 924\).
The absolute difference is given by:
\(|2640 - 924| = 1716\)
Calculate \(12^3 - 12\):
\(12^3 = 1728\)
\(12^3 - 12 = 1728 - 12 = 1716\)
Thus, the absolute difference matches the option \(12^3 - 12\), making it the correct answer.