Question

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

Answer 1

Answer:

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

Step-by-step 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,

${n \choose r}= ^nC_r =\frac{n!}{(n-r)!r!}$

Now by using the above formula.

$\frac{5!}{0!\left(5-0\right)!}\left(2x\right)^5\left(-1\right)^0+\frac{5!}{1!\left(5-1\right)!}\left(2x\right)^4\left(-1\right)^1+\frac{5!}{2!\left(5-2\right)!}\left(2x\right)^3\left(-1\right)^2+\frac{5!}{3!\left(5-3\right)!}\left(2x\right)^2\left(-1\right)^3+\frac{5!}{4!\left(5-4\right)!}\left(2x\right)^1\left(-1\right)^4+\frac{5!}{5!\left(5-5\right)!}\left(2x\right)^0\left(-1\right)^5$

$2^5\cdot \:1\cdot \:1\cdot \:x^5-1\cdot \frac{2^4\cdot \:5x^4}{1!}+1\cdot \frac{2^3\cdot \:20x^3}{2!}-1\cdot \frac{2^2\cdot \:20x^2}{2!}+1\cdot \frac{5\cdot \:2x}{1!}+1\cdot \frac{\left(-1\right)^5}{\left(5-5\right)!}$

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

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