<s>[INST] Solve for $x$ where $x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$ Consider the following equality:
$$x=(1-2)(1+2+4)+(2-3)(4+6+9)+(3-4)(9+12+16)+....+(49-50)(2401+2450+2500)$$
Solve for $x$.
The only thing I noticed is the first part like $(1-2)$,$(3-4)$ gives us $-1$ but then I just don't see what the trick behind this problem is.
 [/INST] Each term is of the form
$$(n-(n+1))(n^2 + (n^2+n) + (n+1)^2) = -(3n^2+3n+1)$$
Hence,
$$x = - \sum_{n=1}^{m} (3n^2+3n+1) = - m(m^2+3m+3)$$
where $m=49$.
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