1706964093

## Answered question

2022-10-07

### Answer & Explanation

alenahelenash

Expert2023-05-29Added 556 answers

To show that the equation
$\frac{1}{\left(1-x{\right)}^{2}}=1+2x+3{x}^{2}+\dots +\left(n+1\right){x}^{n}+\frac{n+2-\left(n+1\right)x}{\left(1-x{\right)}^{2}}{x}^{\left(n+1\right)}$
is true, we'll begin by expanding the right side of the equation and then simplifying it step by step.
Let's start with the expansion:
$1+2x+3{x}^{2}+\dots +\left(n+1\right){x}^{n}+\frac{n+2-\left(n+1\right)x}{\left(1-x{\right)}^{2}}{x}^{\left(n+1\right)}$
To find a common denominator for the terms involving $x$, we can multiply each term by $\left(1-x{\right)}^{2}$:
$\left(1-x{\right)}^{2}+2x\left(1-x{\right)}^{2}+3{x}^{2}\left(1-x{\right)}^{2}+\dots +\left(n+1\right){x}^{n}\left(1-x{\right)}^{2}+\left(n+2-\left(n+1\right)x\right){x}^{\left(n+1\right)}$
Now, let's simplify each term:
$\left(1-x{\right)}^{2}=\left(1-2x+{x}^{2}\right)$
$2x\left(1-x{\right)}^{2}=2x\left(1-2x+{x}^{2}\right)=2x-4{x}^{2}+2{x}^{3}$
$3{x}^{2}\left(1-x{\right)}^{2}=3{x}^{2}\left(1-2x+{x}^{2}\right)=3{x}^{2}-6{x}^{3}+3{x}^{4}$
Continuing this pattern, we can simplify the remaining terms:
$\left(n+1\right){x}^{n}\left(1-x{\right)}^{2}=\left(n+1\right){x}^{n}-2\left(n+1\right){x}^{n+1}+\left(n+1\right){x}^{n+2}$
$\left(n+2-\left(n+1\right)x\right){x}^{\left(n+1\right)}=\left(n+2\right){x}^{\left(n+1\right)}-\left(n+1\right){x}^{\left(n+2\right)}$
Now, let's combine all the simplified terms:
$\left(1-2x+{x}^{2}\right)+\left(2x-4{x}^{2}+2{x}^{3}\right)+\left(3{x}^{2}-6{x}^{3}+3{x}^{4}\right)+\dots +\left(\left(n+1\right){x}^{n}-2\left(n+1\right){x}^{n+1}+\left(n+1\right){x}^{n+2}\right)+\left(\left(n+2\right){x}^{\left(n+1\right)}-\left(n+1\right){x}^{\left(n+2\right)}\right)$
We can see that the common terms cancel each other out, leaving only the outermost terms:
$1+\left(n+2\right){x}^{\left(n+1\right)}-\left(n+1\right){x}^{\left(n+2\right)}$
Now, let's simplify the terms further:
$1+\left(n+2\right){x}^{\left(n+1\right)}-\left(n+1\right){x}^{\left(n+2\right)}=\frac{1}{\left(1-x{\right)}^{2}}+\frac{\left(n+2\right){x}^{\left(n+1\right)}}{\left(1-x{\right)}^{2}}-\frac{\left(n+1\right){x}^{\left(n+2\right)}}{\left(1-x{\right)}^{2}}$
Since all terms have a common denominator, we can combine them into a single fraction:
$\frac{1}{\left(1-x{\right)}^{2}}+\frac{\left(n+2\right){x}^{\left(n+1\right)}}{\left(1-x{\right)}^{2}}-\frac{\left(n+1\right){x}^{\left(n+2\right)}}{\left(1-x{\right)}^{2}}=\frac{1+\left(n+2\right){x}^{\left(n+1\right)}-\left(n+1\right){x}^{\left(n+2\right)}}{\left(1-x{\right)}^{2}}$
Finally, we can see that the numerator of the fraction is the same as the right side of the original equation. Therefore, we have:
$\frac{1}{\left(1-x{\right)}^{2}}=1+2x+3{x}^{2}+\dots +\left(n+1\right){x}^{n}+\frac{n+2-\left(n+1\right)x}{\left(1-x{\right)}^{2}}{x}^{\left(n+1\right)}$
And that concludes the proof.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?