What is the coefficient of the term in #x^9# in the expansion of #(3+x^3)^5# ?
1 Answer
May 28, 2017
Explanation:
The Binomial Theorem tells us that:
#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#
where
So with
#(3+x^3)^5 = sum_(k=0)^5 ((5),(k)) 3^(5-k) (x^3)^k#
The term in
#((5),(color(blue)(3))) 3^(5-color(blue)(3)) (x^3)^color(blue)(3) = (5!)/(2!3!)*9x^9#
#color(white)(((5),(3)) 3^(5-3) (x^3)^3) = (5*4)/2*9x^9#
#color(white)(((5),(3)) 3^(5-3) (x^3)^3) = 90x^9#
So the coeffient of
Instead of calculating