Question

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:

• A. \(10^3 - 10\)
• B. \(12^3 - 12\)
• C. \(11^3 - 11\)
• D. \(13^3 - 13\)

Hint:

When dealing with expansions, simplify the general term first, then identify the required terms using the exponent relationships.

Answer 1

The Correct Option is B

The general term in the expansion of \( \left( 2x^2 + \frac{1}{2x} \right)^{11} \) is:

\[ T_{r+1} = \binom{11}{r} (2x^2)^{11-r} \left( \frac{1}{2x} \right)^r. \]

Simplify:

\[ T_{r+1} = \binom{11}{r} 2^{11-r} x^{2(11-r)} \cdot \frac{1}{2^r x^r}. \]

\[ T_{r+1} = \binom{11}{r} \frac{2^{11-r}}{2^r} x^{22 - 3r}. \]

\[ T_{r+1} = \binom{11}{r} 2^{11-2r} x^{22 - 3r}. \]

For the coefficient of \( x^{10} \):

\[ 22 - 3r = 10 \implies r = 4. \]

The coefficient of \( x^{10} \) is:

\[ \binom{11}{4} 2^{11 - 2(4)} = \binom{11}{4} 2^{3}. \]

For the coefficient of \( x^7 \):

\[ 22 - 3r = 7 \implies r = 5. \]

The coefficient of \( x^7 \) is:

\[ \binom{11}{5} 2^{11 - 2(5)} = \binom{11}{5} 2^{1}. \]

Now calculate the absolute difference:

\[ \text{Difference} = \left| \binom{11}{4} 2^3 - \binom{11}{5} 2^1 \right|. \]

Expand the binomial coefficients:

\[ \binom{11}{4} = \frac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 330, \]

\[ \binom{11}{5} = \frac{11 \cdot 10 \cdot 9 \cdot 8 \cdot 7}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 462. \]

Substitute into the expression:

\[ \text{Difference} = \left| 330 \cdot 8 - 462 \cdot 2 \right|. \]

\[ \text{Difference} = \left| 2640 - 924 \right| = 1716. \]

Finally, express \(1716\) as \( 12^3 - 12 \):

\[ 12^3 = 1728, \quad 12^3 - 12 = 1728 - 12 = 1716. \]