Question

Coefficient of \(x^{10}\) in the expansion of \[ \left(x^2+\frac1x\right)^{12} + \left(x+\frac1{x^2}\right)^{12} \] is:

• A. \(12\)
• B. \(66\)
• C. \(112\)
• D. \(0\)

Hint:

For binomial coefficient problems, first write the exponent of \(x\) in the general term and equate it to the required power.

Answer 1

The Correct Option is D

Step 1: Find coefficient from first expansion.
General term: \[ T_{r+1} = \binom{12}{r} (x^2)^{12-r} \left(\frac1x\right)^r. \] Power of \(x\): \[ 24-3r. \] Set equal to \(10\), \[ 24-3r=10. \] \[ r=\frac{14}{3}. \] Not integer. Hence contribution \(=0\).

Step 2: Find coefficient from second expansion.
General term: \[ T_{r+1} = \binom{12}{r} x^{12-r} \left(\frac1{x^2}\right)^r. \] Power of \(x\): \[ 12-3r. \] Set equal to \(10\), \[ 12-3r=10. \] \[ r=\frac23. \] Not integer. Hence contribution \(=0\). Therefore total coefficient \[ \boxed{0}. \]