Question

What is the second term of `(p+q)^5`?

Answer 1

`5p^4q`

Explanation:

Use the binomial theorem

`(p+q)^n=sum_(k=0)^(n)(n!)/((k!)(n-k)!)p^(n-k)q^k`

For the second term, `n`=5 and `k`=1 (`k` is 1 for the second term and 0 for the first term) so the we compute the term in the summation when `k`=1

`(5!)/((1!)(5-1)!)p^(5-1)q^1=5p^4q`

Because this problem is so short, let's expand the ENTIRE expression to give you a better picture of what's going on.

`(p+q)^5=(5!)/((0!)(5-0)!)p^(5-0)q^0+(5!)/((1!)(5-1)!)p^(5-1)q^1+(5!)/((2!)(5-2)!)p^(5-2)q^2+(5!)/((3!)(5-3)!)p^(5-3)q^3+(5!)/((4!)(5-4)!)p^(5-4)q^4+(5!)/((5!)(5-5)!)p^(5-5)q^5`

`=(5!)/((1)5!)p^5+(5!)/((1)4!)p^4q^1+(5!)/(2!3!)p^3q^2+(5!)/(3!2!)p^(2)q^3+(5!)/(4!(1))p^1q^4+(5!)/(5!(1))q^5`

`=p^5+5p^4q^1+(5*4)/2p^3q^2+(5*4)/2p^(2)q^3+5p^1q^4+q^5`

`=p^5+5p^4q+10p^3q^2+10p^(2)q^3+5pq^4+q^5`