According to a source, the answer is $x^2-\frac{1}{4}$
Please explain
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$\begingroup$Let $n$ be the first of those consecutive numbers. Then we have: $$ n + (n+1) = x $$
which implies $ n= \frac{x-1}{2}$
using that you can calculate the product of $n(n+1)$
$\endgroup$ 4 $\begingroup$Let the consecutive integers be $a$,$a+1$...
Now,$a+(a+1)=x\implies a=\frac{x-1}{2}\implies (a+1)=\frac{x+1}{2}$
Hence, $a(a+1)=\frac{x-1}{2}*\frac{x+1}{2}=\frac{x^2-1}{4}$
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