A probability problem by Megha Malyala

Probability Level 2

Find the 4th term in the expansion ( x − 2 y ) 12 (x-2y)^{12}

1760 x 9 y 3 1760x^{9}y^{3} − 1760 x 9 y 3 -1760x^{9}y^{3} − 1760 x 3 y 6 -1760x^{3}y^{6} 1760 x 3 y 6 1760x^{3}y^{6}

Megha Malyala by Megha Malyala , 30, India

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1 solution

Chew-Seong Cheong
Jul 28, 2017

The fourth term is given by ( 12 3 ) x 12 − 3 ( − 2 y ) 3 = 12 × 11 × 10 2 × 3 x 9 ( − 2 3 y 3 ) = − 1760 x 9 y 3 \displaystyle {12 \choose 3}x^{12-3}(-2y)^3 = \dfrac {12\times 11 \times 10}{2\times 3} x^9 (-2^3y^3) = \boxed{-1760x^9y^3} .

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