How do you find the 6th term of #(a + b)^8# ?

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Answer 1 · 2016-03-19T19:54:37

You must use the formula #t_(r + 1) = nC_r xx a^(n - r) xx b^r#

n is the expression's exponent

#r + 1 = 6#

#r = 5#

#nC_r# is calculated by the formula #(n!)/((n - r)!(r!))#

n = 8 and r = 5. Calculating, we get:

#(8!)/((3! xx 5!)#

= 56

#t_6 =56 xx a^(8 - 5) xx b^5#

#t_6 = 56a^3b^5#

Hopefully this helps!