Constants Rule

Probability Level 2

Find the constant term in the expansion of ( x 2 + 1 x ) 12 . \large \left(x^2 + \frac{1}{x}\right)^{12}.

567 549 564 495

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Hana Wehbi
Jun 9, 2016

According to the binomial theorem: ( x 2 + 1 x ) 12 = ∑ k = 0 12 ( 12 k ) ( x 2 ) k ( 1 x ) 12 − k \Large(x^{2}+ \frac{1}{x})^{12} = \sum_{k=0}^{12}{12\choose k}(x^{2})^{k}(\frac{1}{x})^{12-k}

( x 2 + 1 x ) 12 = ∑ k = 0 12 ( 12 k ) ( x ) 3 k − 12 \Large(x^{2}+ \frac{1}{x})^{12}= \sum_{k=0}^{12}{12\choose k}(x)^{3k-12} ;

Set: 3 k − 12 = 0 ⟹ k = 4 \large 3k-12=0 \implies k=4 ;

Thus, the constant term is ( 12 4 ) = 495 \Large {12\choose 4} = \boxed{495}

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