alleryThe constant term in the expansion of $\left(x^2+\dfrac{1}{x}\right)^6$ is:
A6
B1
C15
D20
Answer & Solution
Correct answer: C. 15
T_{r+1}=C(6,r)x^{2(6−r)}x^{−r}=x^{12−3r}; constant ⟹ 12−3r=0 ⟹ r=4; C(6,4)=15.
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