What is the coefficient of the term in #x^9# in the expansion of #(3+x^3)^5# ?

1 Answer
May 28, 2017

#((5),(3)) 3^2 = 90#

Explanation:

The Binomial Theorem tells us that:

#(a+b)^n = sum_(k=0)^n ((n),(k)) a^(n-k) b^k#

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

So with #n=5#, #a=3# and #b=x^3# we find:

#(3+x^3)^5 = sum_(k=0)^5 ((5),(k)) 3^(5-k) (x^3)^k#

The term in #x^9# occurs when #k=3# giving us:

#((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 #x^9# is #90#

Instead of calculating #((5),(3)) = 10# directly, we can read it from Pascal's triangle, in the row starting #1, 5,...#

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