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

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Answer 1 · 2016-07-19T11:51:30

Coefficient is 405.

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

where $((n),(k)) = (n!)/(k!(n-k)!)$

For $n=5$ and $y=3$ we are looking for the coefficient of $x^1$ . This means we need $n-k = 1 implies k = 4$ .

$((5),(4))x^1*3^4 = (5!)/(4!1!)*81*x = 405x$

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