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 · 2016-08-13T17:55:42

80

Binomial theorem:

$(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k)y^k$

$(x+2)^5 = sum_(k=0)^5((5),(k)) x^(5-k)2^k$

Looking for $x^2$ so look at the $k=3$ term:

$((5),(3))x^2*2^3 = 8*(5!)/(3!2!)x^2 = 80x^2$

How do you find the coefficient of x^2x2x^2 in the expansion of (2+x)^5(2+x)5(2+x)^5?