Interesting arithmetic sequence

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SUMMARY

The discussion centers on proving that for the arithmetic sequence defined by N = 1.2.3 + 2.3.4 + ... + n(n+1)(n+2), the expression 4N + 1 results in a perfect square for positive integers n. The formula for the sum of the sequence is established as (n+1)n(n-1)(n-2)/4. The key simplification involves demonstrating that (n+1)n(n-1)(n-2) + 1 is a square, which is a more manageable problem to solve.

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Given N= 1.2.3 + 2.3.4 + ... + n(n+1)(n+2), prove that 4N + 1 is a square (n is a positive integer)
 
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It's easy to see that (see www.math.uic.edu/~kauffman/DCalc.pdf on how to find such things)

[tex]1.2.3+...+n(n-1)(n-2)=\frac{(n+1)n(n-1)(n-2)}{4}[/tex]

Thus it suffices to show that

[tex](n+1)n(n-1)(n-2)+1[/tex]

is a square. Which is a much simpler problem.
 
Great!
 

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