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

1 Answer

Answer:

#(C_(8,3))a^5b^3=(8xx7xx6xx5!)/(5!xx3xx2)a^5b^3=color(blue)(ul(bar(abs(color(black)(56a^5b^3#

Explanation:

The structure of the terms in a binomial expansion follows:

#(a+b)^n=(C_(n,0))a^nb^0+(C_(n,1))a^(n-1)b^1+...+(C_(n,n))a^0b^n#

We're being asked to find the 4th term of #(a+b)^8#. That is equal to:

#(C_(8,3))a^5b^3=(8xx7xx6xx5!)/(5!xx3xx2)a^5b^3=color(blue)(ul(bar(abs(color(black)(56a^5b^3#

Keep in mind that we start counting the terms from 0 (and so there are 9 terms in total with the exponent equal to 8). This means the 4th term up from 0 is 3 (0, 1, 2, 3) and from 8 is 5 (8, 7, 6, 5).