Question

The 1st term in the binomial (2v + 4u)^4?

The answer should be 16v^4 but I have no idea how to get there

Answer 1

See below.

Explanation:

For binomial expansions of the form:

`(x+y)^n`

We have:

`sum_(r=0)^(n)((n),(r))x^(n-r)y^r`

Where:

`((n),(r))=color(white)(0)^nC_(r)=(n!)/((r!(n-r)!)`

In this expansion the powers of `x` are decreasing, so we would expect the power of `x` in the first term to be `n`.

`:.`

For:

`x^(n-r)=x^n=>r=0`

For given expansion:

`(2v+4u)^4`

Using the above:

`sum_(r=0)^(4)((4),(0))(2v)^(4-0)(4u)^0`

`sum_(r=0)^(4)((4),(0))(2v)^(4)` ( any number to the zero power is 1 )

`((4),(0))(2v)^4=((4),(0))16v^4`

`((4),(0))=color(white)(0)^4C_(0)=(4!)/(0!(4-0)!)=(4xx3xx2xx1)/(1xx4xx3xx2xx1)`

`=(cancel(4)xxcancel(3)xxcancel(2)xxcancel(1))/(1xxcancel(4)xxcancel(3)xxcancel(2)xxcancel(1))=1/1=1`

`((4),(0))16v^4=1xx16v^4=16v^4`

Note:

`0! =1`