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

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Answer 1 · 2018-07-03T07:36:48

The answer is #=720#

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

Where,

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

Here, we have

#(2x+3)^5#

And we need the coefficient of #x^3#

#n=5#

#a=2x#

#b=3#

#k=2#

The coefficient is

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

#=(5!)/((3!)(2!))* 8* 9#

#=10*72#

#=720#