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

1 Answer
kceylan
Oct 30, 2016

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 +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 +5.2^4x+ 10.2^3x^2+10.2^2x^3+5.2x^4+x^5#
or #(2+x)^5=32 +80x+ 80x^2+40x^3+10x^4+x^5# Therefore the coefficient of #x^2 # in the expansion of #(2+x)^5# is 80.

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