How do you find the coefficient of $x^{2}$ $x^2$ in the expansion of ${(2 + x)}^{5}$ $(2+x)^5$ ?

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How do you find the coefficient of $x^{2}$ $x^2$ in the expansion of ${(2 + x)}^{5}$ $(2+x)^5$ ?

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Answer 1 · 2016-10-30T18:19:49

80

Explanation:

Binomial coefficient are may be found by applyıng the Pascal's rule . Pascal's rule or the Pascal's triangle can easiliy be found in the internet. ${(x + y)}^{5} = x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5}$ $(x+y)^5=x^5 +5x^4y+ 10x^3y^2+10x^2y^3+5xy^4+y^5$
the x in the question correspond to y in the formulation. Acoording to this ${(2 + x)}^{5} = 2^{5} + {5.2}^{4} x + {10.2}^{3} x^{2} + {10.2}^{2} x^{3} + 5.2 x^{4} + x^{5}$ $(2+x)^5=2^5 +5.2^4x+ 10.2^3x^2+10.2^2x^3+5.2x^4+x^5$
or ${(2 + x)}^{5} = 32 + 80 x + 80 x^{2} + 40 x^{3} + 10 x^{4} + x^{5}$ $(2+x)^5=32 +80x+ 80x^2+40x^3+10x^4+x^5$ Therefore the coefficient of $x^{2}$ $x^2$ in the expansion of ${(2 + x)}^{5}$ $(2+x)^5$ is 80.

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How do you find the coefficient of $x^{2}$ $x^2$ in the expansion of ${(2 + x)}^{5}$ $(2+x)^5$ ?

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How do you find the coefficient of $x^{2}$ $x^2$ in the expansion of ${(2 + x)}^{5}$ $(2+x)^5$ ?