How do you find the 5th term in the binomial expansion for #(5a + 6b)^5#?

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Answer 1 · 2015-08-31T10:13:15

Use general formula for binomial expansion and evaluate 5th term as:

#32400ab^4#

In general, #(A+B)^N = sum_(n=0)^N ((N),(n)) A^(N-n) B^n#

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

So, with #A=5a#, #B=6b# and #N=5# we get:

#(5a+6b)^5 = sum_(n=0)^5 ((5),(n)) (5a)^(5-n) (6b)^n#

where #((5),(n)) = (5!)/(n! (5-n)!)#

The 5th term is the one for #n=4#, that is:

#((5),(4)) (5a)^(5-4) (6b)^4#

#=5*(5a)^1*(6b)^4#

#=5*5a*6^4*b^4#

#=5*5a*1296*b^4#

#=32400ab^4#