Question

Use the binomial theorem to compute (2x-1)^5

Answer 1

The expended form of the provided expression is:
`32x^5-80x^4+80x^3-40x^2+10x-1`

Explanation:

Consider the provided expression.

`(2x-1)^5`

The binomial theorem:

`(a+b)^(n)=\sum _(r=0)^(n){n \choose r}a^(n-r)b^r`

Where,

`{n \choose r}= ^nC_r =(n!)/((n-r)!r!)`

Now by using the above formula.

`(5!)/(0!\left(5-0\right)!)\left(2x\right)^5\left(-1\right)^0+(5!)/(1!\left(5-1\right)!)\left(2x\right)^4\left(-1\right)^1+(5!)/(2!\left(5-2\right)!)\left(2x\right)^3\left(-1\right)^2+(5!)/(3!\left(5-3\right)!)\left(2x\right)^2\left(-1\right)^3+(5!)/(4!\left(5-4\right)!)\left(2x\right)^1\left(-1\right)^4+(5!)/(5!\left(5-5\right)!)\left(2x\right)^0\left(-1\right)^5`

`2^5\cdot \:1\cdot \:1\cdot \:x^5-1\cdot (2^4\cdot \:5x^4)/(1!)+1\cdot (2^3\cdot \:20x^3)/(2!)-1\cdot (2^2\cdot \:20x^2)/(2!)+1\cdot (5\cdot \:2x)/(1!)+1\cdot (\left(-1\right)^5)/(\left(5-5\right)!)`

`32x^5-80x^4+80x^3-40x^2+10x-1`

Hence, the expended form of the provided expression is:
`32x^5-80x^4+80x^3-40x^2+10x-1`