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: Recall the binomial general term.
For $(p+q)^{12}$ the general term is $\binom{12}{r}p^{12-r}q^{r}$. We will track only the power of $x$ each part produces.
Step 2: Handle the first bracket.
For $\left(x^2+\frac1x\right)^{12}$ the term is $\binom{12}{r}(x^2)^{12-r}(x^{-1})^{r}$, whose $x$-power is $24-2r-r=24-3r$.
Step 3: Demand power $10$ from the first bracket.
Set $24-3r=10$, giving $3r=14$, so $r=\frac{14}{3}$, which is not a whole number. Hence no valid term, contribution $0$.
Step 4: Handle the second bracket.
For $\left(x+\frac1{x^2}\right)^{12}$ the term is $\binom{12}{r}x^{12-r}(x^{-2})^{r}$, whose $x$-power is $12-r-2r=12-3r$.
Step 5: Demand power $10$ from the second bracket.
Set $12-3r=10$, giving $3r=2$, so $r=\frac23$, again not a whole number. Contribution $0$.
Step 6: Add the contributions.
Both brackets give $0$, so the coefficient of $x^{10}$ in the sum is $0$.
\[ \boxed{0} \]