 1706964093

2022-10-07 alenahelenash

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.
$1+2x+3{x}^{2}+\dots +\left(n+1\right){x}^{n}+\frac{n+2-\left(n+1\right)x}{\left(1-x{\right)}^{2}}{x}^{\left(n+1\right)}$
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.

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