How Does ((n+1)^2 * n!) / ((n+1)! * n^2) Simplify to (n+1) / n^2?

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The expression ((n+1)^2 * n!) / ((n+1)! * n^2) simplifies to (n+1) / n^2 through a series of algebraic steps. Initially, the factorial (n+1)! is expanded to (n+1) * n!, allowing for cancellation of n!. This results in the expression simplifying to (n+1) * n! / (n! * n^2). After canceling n!, the final simplified form is (n+1) / n^2. The key to the simplification lies in recognizing the factorial expansion.
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How can ((n+1)^2(*n!))/((n+1)!*n^2) be simplified to (n+1)/n^2?

My own answer is (n+1)^2/n^2, but its apparently wrong
 
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Here's how:

<br /> \frac{(n+1)^{2}\cdot n!}{(n+1)!\cdot n^{2}}<br /> =<br /> \frac{(n+1)^{2}\cdot n!}{(n)!\cdot(n+1) \cdot n^{2}}<br /> =<br /> \frac{(n+1)\cdot n!}{(n)! \cdot n^{2}}<br /> =<br /> \frac{(n+1)}{n^{2}}<br /> <br />

All you had to do was expand the (n+1)! into (n+1)n!.

:)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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