How do you find the coefficient of $x^{2}$ $x^2$ in the expansion of ${(x + 3)}^{5}$ $(x+3)^5$ ?

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Answer 1 · 2017-02-27T09:30:51

Coefficient of $x^{2}$ $x^2$ is $270$ $270$

Binomial expansion of ${(x + a)}^{n}$ $(x+a)^n$ is

$Sigma_(r=0)^(r=n)((n),(r))x^(n-r)a^r$ , where $((n),(r))=(n!)/((n-r)!r!)$

Here $n=5$ and as we seek coefficient of $x^2$ ,

$n-r$ i.e. $5-r=2$ or $r=3$ and $a=3$

Hence coefficient of $x^2$ is

$((n),(r))a^r=((5),(3))3^3=(5!)/(2!3!)27=(5xx4xx3xx2xx1)/(2xx1xx3xx2xx1)xx27=270$

How do you find the coefficient of x^2x2x^2 in the expansion of (x+3)^5(x+3)5(x+3)^5?